Problems #11-25 (and a few more)

**Problem #11:** 26.40 mL of a 0.550 M solution of H_{2}SO_{4} is allowed to react with 19.60 mL of a 0.720 M NaOH solution. What is the pH of the new solution after reaction has ceased? (Assume that both hydrogens of the H_{2}SO_{4} dissociate 100%.)

**Solution:**

1) Initial amounts of the reactants:

H_{2}SO_{4}---> (0.550) (26.40) = 14.52 mmol

NaOH ---> (0.720) (19.60) = 14.112 mmol

2) The balanced chemical equation tells us that it takes 2 mmoles of NaOH to react with 1 mmole of H_{2}SO_{4}.

mmol H _{2}SO_{4}+ 2NaOH ---> H _{2}O+ Na _{2}SO_{4}initial 14.52 14.112 0 0 change −7.056 −14.112 14.112 14.112 final 7.464 0 14.112 14.112

3) Determine the molarity of the hydrogen ion, then the pH:

After the reaction there are 7.464 mmoles of H_{2}SO_{4}remaining that did not react.(7.464 mmol H

_{2}SO_{4}) (2 mmol H^{+}/ 1 mmo H_{2}SO_{4}) = 14.928 mmol H^{+}The volume of the solution after the reaction:

26.40 mL + 19.60 mL = 46.00 mL[H

^{+}] = 14.928 mmol / 46.00 mL = 0.3245217 MpH = −log [H

^{+}] = −log 0.3245217 = 0.489

**Problem #12:** What is the final pH of the solution obtained by mixing 0.20 L of 0.15 M HCl and 0.20 L of 0.30 M NaOH?

**Solution:**

1) Determine moles of each solute:

HCl ---> (0.15 mol/L) (0.20 L) = 0.030 mol

NaOH ---> (0.30 mol/L) (0.20 L = 0.060 mol

2) HCl and NaOH react in a 1:1 molar ratio. HCl is seen to be the limiting reagent. How much NaOH remains after the reaction ceases?

0.060 mol − 0.030 mol = 0.030 mol

3) What is the new molarity of the NaOH?

0.030 mol / 0.40 L = 0.075 M

4) Determine the pOH (remember, NaOH is a strong base):

pOH = −log 0.075 = 1.12

5) Determine the pH:

pH + pOH = pK_{w}pH = 14 − 1.12 = 12.88

**Problem #13:** What is the final pH of the solution obtained by mixing 0.20 L of 0.30 M HCl and 0.20 L of 0.15 M NaOH?

**Solution:**

1) Determine moles of each solute:

HCl ---> (0.30 mol/L) (0.20 L) = 0.060 mol

NaOH ---> (0.15 mol/L) (0.20 L = 0.030 mol

2) HCl and NaOH react in a 1:1 molar ratio. NaOH is seen to be the limiting reagent. How much HCl remains after the reaction ceases?

0.060 mol − 0.030 mol = 0.030 mol

3) What is the new molarity of the HCl?

0.030 mol / 0.40 L = 0.075 M

4) Determine the pH (remember, HCl is a strong acid):

pH = −log 0.075 = 1.12

**Problem #14:** 50.0 mL of 1.50 M NaOH is titrated against 2.00 M HCl. What is the pH of the reaction system after addition of 30.0 mL HCl?

**Solution:**

NaOH ---> (1.50 mmol/mL) (50.0 mL) = 75.0 mmol

HCl ---> (2.00 mmol/mL) (30.0 mL) = 60.0 mmolNaOH + HCl ---> NaCl + H

_{2}O75.0 − 60.0 = 15.0 mmol of NaOH remains after reaction

15.0 mmol / 80.0 mL = 0.1875 M

pOH = −log 0.1875 = 0.727

pH = 13.273

**Problem #15:** What is the pH at each of the points in the titration of 25.0 mL of 0.150 M HCl by 0.150 M NaOH:

(a) After adding 25.0 mL NaOH

(b) After adding 26.0 mL NaOH

**Solution:**

1) Discussion about (a):

Note that the volumes and the molarities are equal. Since moles equals molarity times volume, we can see that the moles of HCl and of NaOH are the same. This means all the acid and all the base are neutralized, leaving a solution of only NaCl (the salt of a strong acid and a strong base). The pH of this solution is 7.000.

2) Discussion about (b):

Note that there is one milliliter of NaOH left unreacted. Since the molarities are equal, we can use M_{1}V_{1}= M_{2}V_{2}to solve for the new molarity of NaOH:(0.150 mol/L) (1.00 mL) = (x) (51.0 mL)x = 0.0029412 M (keeping two guard digits)

Since the base was in excess, we solve for the pOH first:

pOH = −log 0.0029412 = 2.531And now, for the pH:

pH = 14.000 − 2.531 = 11.469

**Problem #16:** What is the pH of the solution obtained by mixing 30.0 mL of 0.250 M HCl and 30.0 mL of 0.125 M NaOH?

**Solution:**

1) Moles of each reactant:

HCl ---> (0.250 mol/L) (0.0300 L) = 0.00750 mol

NaOH ---> (0.125 mol/L) (0.0300 L) = 0.00375 molNote that the volumes are equal while the molarity of the NaOH is half that of the HCl.

2) Amount of excess reagent:

0.00750 mol minus 0.00375 mol = 0.00375 mol of HCl

3) New molarity and pH:

0.00375 mol / 0.0600 L = 0.0625 MpH = −log 0.0625 = 1.204

Comment #1: the usual assumption in these types of problems is that the volumes are additive. Strictly speaking, this is not true. However, the volume difference after adding is so tiny that we can, at any but the most exact level, completely ignore any "non-additive volume" issues. Sometimes, the problem will instruct you to assume the volumes are additive. If that instruction is not present, your best choice is to assume that it is present.

Comment #2: you could use M_{1}V_{1} = M_{2}V_{2} to solve this problem.

Notice that 0.250 M is exactly double 0.125 M. Notice also that the volumes are exactly the same. This means the acid is double the number of moles as compared to the base. So, the 0.125 M of the base neutralizes exactly half of the acid, turing it from 0.250 M to 0.125 M.

However, in so doing, the volume went from 30 mL to 60 mL. So:

(0.125 M) (30.0 mL) = (x) (60.0 mL)

x = 0.0625 M

Advice: if you see this method can be used on a test question, you should probably go the traditional route in your answer. Your teacher might consider your method to not be worth full points. If you have time, add in the method in comment #2. Perhaps some bonus points!

**Problem #17:** 19.8 L of 2.31 M hydrobromic acid were mixed with 80.2 L of sodium hydroxide to produce a solution having a pH of 3.11. What was the concentration of the base?

**Solution:**

HBr + NaOH ---> NaBr + H_{2}OThere is a 1:1 molar ratio between HBr and NaOH

Moles of acid before reaction:

(2.31 mol/L) (19.8 L) = 45.738 molFrom pH of solution, get [H

^{+}] after reaction has ceased:10^{-3.11}= 7.7625 x 10^{-4}MDetermine total moles of H

^{+}in solution after reaction has ceased:(7.7625 x 10^{-4}) (100 L) = 0.077625 mol100 L comes from 19.8 + 80.2

This many moles of H

^{+}were neutralized by hydroxide:45.738 mol − 0.077625 mol = 45.660375 molConcentration of NaOH solution that was used:

45.660375 mol / 80.2 L = 0.569 M

**Problem #18:** What is the pH of the solution that results from mixing 350. mL of 0.600M HBr with 225 mL of 0.500 M KOH?

**Solution:**

moles HBr ---> (0.600 mol/L) (0.350 L) = 0.210 mol

moles KOH ---> (0.500 mol/L) (0.225 L) = 0.1125 molHBr + KOH ---> KBr + H

_{2}OHBr and KOH react in 1:1 molar ratio. KOH is the limiting reagent.

amount HBr left ---> 0.210 minus 0.1125 = 0.0975 mol

new molarity of HBr ---> 0.0975 mol / 0.575 L = 0.170 M (to three sig figs)

HBr is a strong acid, so it ionizes 100%. This means the [H

^{+}] in the solution is 0.170 MpH = −log 0.170 = 0.770

**Problem #19:** What is the pH of solution formed by mixing 140.0 mL of 0.0150 M HCl with 95.0 mL of 0.0550 M of NaOH?

**Solution:**

1) Calculate moles of HCl and NaOH:

moles HCl: (0.0150 mol/L) (0.1400 L) = 0.00210 mol

moles NaOH: (0.0550 mol/L) (0.095 L) = 0.005225 mol

2) The HCl and NaOH react in a 1:1 ratio. HCl runs out first, so:

0.005225 mol minus 0.00210 mol = 0.003125 mol of NaOH remaining

3) Calculate new molarity:

0.003125 mol divided by 0.235 L = 0.013298 M (I'll keep a couple guard digits.)

4) Calculate pOH:

pOH = - log 0.013298 = 1.876

5) Calculate pH:

14.000 minus 1.876 = 12.124

**Problem #20:** What is the pH of a 0.010 M solution of HNO_{3} mixed in equal amounts with a 0.13 M solution of KOH?

**Solution:**

1) I will use 1.00 L of each solution. You can use any amount, they just have to be equal. The first thing to do is determine the moles of each substance:

HNO_{3}---> (0.010 mol/L) (1.00 L) = 0.010 mol

KOH ---> (0.13 mol/L) (1.00 L) = 0.13 mol

2) The acid and the base react in a 1:1 molar ratio and the KOH is waaaaaay in excess, so we determine how much KOH remains:

0.13 mol − 0.01 mol = 0.12 mol (of KOH remains)

3) In order to calculate the pH, we need the molarity. However, keep in mind we used 1.00 L of KOH solution and 1.00 L of HNO_{3} solution, so the new volume is 2.00 L.

0.12 mol / 2.00 L = 0.060 M

4) Next, we calcuate the pOH:

pOH = −log 0.060 = 1.22

5) Now, for the pH:

pH + pOH = 14

pH = 14 − 1.22 = 12.78

**Problem #21:** When 53.0 mL of 0.320 M KOH is mixed with 28.0 mL of 0.250 M HNO_{3}, what is the concentration of the OH¯ ion after the reaction goes to completion? What is the pH?

**Solution:**

moles OH¯ from KOH ---> (0.320 mol/L) (0.0530 L) = 0.01696 molmoles H

^{+}from HNO_{3}---> (0.250 mol/L) (0.0280 L) = 0.00700 molThe H

^{+}and OH¯ react in a 1:1 ratio. The OH¯ is in excess.0.01696 mol − 0.00700 mol = 0.00996 mol <--- hydroxide left over

0.00996 mol / 0.0810 L = 0.122963 M

To three sig figs, 0.123 M

pOH = −log 0.122963 = 0.91022

pH = 14 − 0.91022 = 13.08978

To three sig figs, 13.090 (and yes, that is three sig figs. It's a logarithm.)

**Problem #22:** A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 16.0 mL of KOH.

**Solution:**

1) We need to determine the moles of HBr and of KOH.

HBr ---> (0.15 mol/L) (0.0500 L) = 0.0075 mol

KOH ---> (0.25 mol/L) (0.0160 L) = 0.0040 mol

2) The KOH and HBr react in a 1:1 molar ratio. To determine how much HBr remains, we subtract:

0.0075 mol − 0.0040 mol = 0.0035 mol

3) The total volume of the two combined solutions is 65.0 mL. The next step is to calculate the new molarity:

0.0035 mol / 0.0650 L = 0.053846 M <--- I'll keep a few extra digits for the pH calculation

4) Since HBr is a strong acid, we know it is 100% ionized in solution, so the H^{+} concentration is 0.053846 M. Now, we calculate the pH:

pH = −log 0.053846 = 1.2688Two sig figs seems best, so 1.27 <--- and yes, that is two sig figs. It's a logarithm.

**Problem #23:** Calculate the molar mass of an unknown acid, H_{2}A, if a 12.11 g sample of the acid takes 38.38 mL of 5.000 M NaOH to neutralize it.

**Solution:**

1) Write the balanced chemical equation:

H_{2}A + 2NaOH ---> Na_{2}A + 2H_{2}OThe key point is that two moles of NaOH are required to neutralize one mole of H

_{2}A.

2) Determine moles of NaOH:

(5.000 mol/L) (0.03838 L) = 0.1919 mol

3) Determine moles of H_{2}A that reacted:

0.1919 mol / 2 = 0.09595 molRemember, two moles of NaOH are consumed for every one mole of the acid. Twice as many moles of NaOH are used as compared to moles of H

_{2}A.

4) Determine the molar mass of H_{2}A:

12.11 g / 0.09595 mol = 126.2 g/mol

**Problem #24:** You have 35.80 mL of a 0.0155 M solution of H_{2}A (a generic acid that is strong in both its hydrogens). You mix in 0.741 g of solid Ca(OH)_{2}. Assuming no volume change, what is the pH of the resulting solution?

**Solution:**

1) Let's get the amounts of hydrogen ion and hydroxide ion:

(0.0155 mmol/mL) (35.80 mL) (2) = 1.1098 mmol of H^{+}(0.741 g / 74.0918 g/mol) (2) = 0.0200 mmol of OH¯

2) H^{+} and OH¯ react in a 1:1 molar ratio:

H^{+}+ OH¯ ---> H_{2}O

3) Hydrogen ion is clearly in excess. How much remains after reaction ceases?

1.1098 mmol − 0.0200 mmol = 1.0898 mmol

4) Calculate the new hydrogen ion concentration:

1.0898 mmol / 35.80 mL = 0.03044134 M

5) Determine the pH:

pH = −log 0.03044134 = 1.516

**Problem #25:** 23.0 g of HCl and 28.0 g of NaOH are added to 500. mL of solution. What is the final pH?

**Solution:**

1) Determine moles of each reactant:

HCl ---> 23.0 g / 36.4609 g/mol = 0.630813 mol

NaOH ---> 28.0 g / 39.9969 g/mol = 0.700054 mol

2) HCl and NaOH react in a 1:1 molar ratio. Determine the excess amount left after reaction:

0.700054 mol − 0.630813 mol = 0.069241 mol of NaOH remains

3) Determine molarity of the NaOH:

0.069241 mol / 0.500 L = 0.138482 M

4) Determine pOH, then pH:

pOH = −log 0.138482 = 0.859pH = 14 − 0.859 = 13.141

**Problem #26:** What volume of HCl solution with a pH of 2.12 can be neutralized by 500. mg of Mg(OH)_{2}?

**Solution:**

1) Determine millimoles of Mg(OH)_{2}:

500. mg / 58.3188 mg/mmol = 8.57356 mmol

2) Determine mmol of HCl required to neutralize:

2HCl + Mg(OH)_{2}---> MgCl_{2}+ 2H_{2}O8.57356 mmol of Mg(OH)

_{2}requires 17.14712 mmol of HCl for complete neutralization.

3) Determine molarity of HCl solution:

[H^{+}] = 10¯^{pH}= 10¯^{2.12}= 0.00758578 M

4) Determine required volume of HCl solution to deliver 17.14712 mmol:

17.14712 mmol / 0.00758578 mmol/mL = 2260 mL

**Problem #27:** A solution is prepared by mixing 29.0 g of H_{2}SO_{4} with 45.0 g of KOH. What is the pH of the solution if the volume is 550. mL?

**Solution:**

1) Write the balanced chemical equation:

H_{2}SO_{4}+ 2KOH ---> K_{2}SO_{4}+ 2H_{2}OThe key point is that, for every one mole of H

_{2}SO_{4}consumed, 2 moles of KOH are required.

2) Determine moles of the two substances involved in the reaction:

29.0 g / 98.0791 g/mol = 0.29568 mol H_{2}SO_{4}

45.0 g / 56.1056 g/mol = 0.80206 mol KOH

3) Determine excess reagent:

0.29568 mole of H_{2}SO_{4}would react completely with 0.29568 x 2 = 0.59136 mole of KOH, but there is more KOH present than that, so KOH is in excess.

4) Determine amount of KOH remaining after reaction:

(0.80206 mol KOH initially) − (0.59136 mol KOH reacted) = 0.2107 mol KOH excess

5) KOH is a strong base, so it ionizes 100% in solution. Determine the molarity of the hydroxide ion:

0.2107 mol KOH produces 0.2107 mol hydroxide ion0.2107 mol / 0.550 L = 0.38309 M in OH¯

6) Determine the pOH, then the pH:

pOH = −log 0.38309 = 0.4167pH = 14 − pOH = 13.583

**Problem #28:** If 3.00 g NaOH is added to 500. ml of 0.10 M HCl, will the resulting solution be acidic or basic?

**Solution:**

1) Determine molar amounts:

NaOH ---> 3.00 g / 40.0 g/mol = 0.075 mol

HCl ---> (0.10 mol/L) (0.5 L) = 0.050 mol

2) Which is the limiting reagent?

HCl and NaOH react in a 1:1 molar ratio, as seen in this equation:HCl + NaOH ---> NaCl + H_{2}O0.05 mole of HCl will react with 0.05 mole of NaOH, leaving 0.025 mole of NaOH unreacted.

3) Is the resulting solution acid or basic?

Basic.

**Problem #29:** A solution is made by adding 0.400 g of solid Ca(OH)_{2}, 40.0 mL of 1.05 M HNO_{3}, and enough water to make a final volume of 75.0 mL. What is the pH?

**Solution:**

1) We need to find out if the Ca(OH)_{2} or the HNO_{3} is in excess and how much of the excess reagent remains.

moles Ca(OH)_{2}---> 0.400 g / 74.0918 g/mol = 0.0053987 mol

moles HNO_{3}---> (1.05 mol/L) (0.0400 L) = 0.0420 mol

2) The chemical reaction is this:

2HNO_{3}+ Ca(OH)_{2}---> Ca(NO_{3})_{2}+ 2H_{2}O

3) We can determine the excess by inspection. Since 2 moles of HNO_{3} are required for one mole of Ca(OH)_{2}, we determine the moles of HNO_{3} required for reacting completely with the Ca(OH)_{2} to be:

0.0053987 mol x 2 = 0.0107974 mol

4) The moles of HNO_{3} remaining after reaction are:

0.0420 − 0.0107974 = 0.0312026 mol

5) We need the molarity of the acid:

0.0312026 mol / 0.0750 L = 0.416035 M

6) The pH is this:

−log 0.416035 = 0.381

Note that HNO_{3} is a strong acid, so it ionizes 100% in solution.

**Problem #30:** What quantity of NaOH(s) must be added to 1.00 L of 0.400 M HCl to achieve a pH of 11.000? (Assume no volume change.)

**Solution:**

moles HCl ---> (0.400 mol/L) (1.00 L) = 0.400 mol

HCl + NaOH ---> NaCl + H

_{2}OThere is a 1:1 molar ratio between HCl and NaOH. Therefore:

Adding 0.400 mole of NaOH(s) will neutralize all the HCl, making a solution with pH = 7.000

To make it pH = 11.000, we consider the pOH when the pH = 11.000

pH + pOH = pK

_{w}11.000 + x = 14.000

pOH = 3.000

[OH¯] = 10¯

^{pOH}= 10¯^{3.000}= 0.00100 Mmoles NaOH ---> (0.00100 mol/L) (1.00 L) = 0.00100 mol

We need to add 0.401 moles of NaOH to achieve a pH of 11.000.

**Problem #31:** How much KOH must be dissolved in 1.00 L of solution so that 25.0 mL of this solution is required to titrate to the equivalence point a solution containing 0.322 g of a monoprotic acid with molar mass 204.6 g/mol?

**Solution:**

1) Determine moles of acid:

0.322 g / 204.6 g/mol = 0.0015738 mol

2) Determine moles of base required in 1.00 L:

0.0015738 mol KOH required in 25.0 mL0.0015738 mol x 40 = 0.062952 mol KOH required in 1.00 L

The 40 came from 1000 / 25. There are forty 25.0 mL amounts in 1000 mL.

3) Determine mass of KOH required:

(0.062952 mol) (56.1049 g/mol) = 3.53 g (to three sig figs)

**Problem #32:** After 10.00 gram sample containing Na_{2}CO_{3} and inert substances was treated with 41.04 mL of 3.000 M HCl, it took 4.22 mL of 1.000 M NaOH to neutralize the excess HCl. Calculate the percent of Na_{2}CO_{3} present in the sample.

**Solution:**

1) We need to determine the excess acid that was used. The equation to do that is this:

M _{a}V_{a}M _{b}V_{b}––––– = ––––– n _{a}n _{b}

(3.000 mol/L) (V _{a})(1.000 mol/L) (4.22 mL) ––––––––––––––– = ––––––––––––––––––– 1 1 V

_{a}= 1.4067 mL

2) Determine the moles of acid used to react with the sodium carbonate:

41.04 mL − 1.4067 mL = 39.6333 mL(3.000 mol/L) (0.0396333 L) = 0.1188999 mol

3) Write the chemical reaction beween HCl and Na_{2}CO_{3}:

2HCl + Na_{2}CO_{3}---> 2NaCl + CO_{2}+ H_{2}OIt requires two moles of HCl to react with one mole of sodium carbonate.

4) Determine moles of sodium carbonate that reacted:

0.1188999 mol is to 2 as x is to 1x = 0.05944995 mol

5) Determine mass of sodium carbonate:

(0.05944995 mol) (105.9888 g/mol) = 6.30 g

6) Determine percent Na_{2}CO_{3} in the sample:

(6.30 / 10.00) * 100 = 63.0%