Problems #1 - 15

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I answered this Yahoo Answers question:

"When H_{2}SO_{4}and NaOH is reacted, is it the H^{+}and OH¯ that has equal number of moles, or the H_{2}SO_{4}and NaOH has equal number of moles?"

I think it's a good question and I wanted to show you the question first, if you wanted to ponder it before looking at my answer.

**Problem #1:** A volume of 20.00 mL of 0.250 M Al(OH)_{3} neutralizes a 75.00 mL sample of H_{2}SO_{4} solution. What is the concentration of H_{2}SO_{4}?

**Solution:**

2Al(OH)_{3}+ 3H_{2}SO_{4}---> Al_{2}(SO_{4})_{3}+ 6H_{2}OAl(OH)

_{3}and H_{2}SO_{4}are in a 2 : 3 molar ratio.moles Al(OH)

_{3}---> (0.250 mol/L) (20.00 mL) = 5.00 millimoles2 is to 3 as 5.00 is to x

x = 7.50 millimoles

molarity H

_{2}SO_{4}---> 7.50 millimoles / 75.00 mL = 0.100 M

**Problem #2:** A 21.62 mL sample of Ca(OH)_{2} solution was titrated with 0.2545 M HCl. 45.87 mL of the acid was required to reach the endpoint of the titration. What was the molarity of calcium hydroxide solution?

**Solution:**

2HCl + Ca(OH)_{2}---> CaCl_{2}+ 2H_{2}Omoles HCl ---> (0.2545 mol/L) (0.04587 L) = 0.011674 mol

use the 2 : 1 molar ratio of HCl to Ca(OH)

_{2}:2 is to 1 as 0.011674 mol is to xx = 0.005837 mol of Ca(OH)

_{2}molarity of Ca(OH)

_{2}---> 0.005837 mol / 0.02162 L = 0.2700 M (to four sig figs)

**Problem #3:** Calculate the volume of NaOH necessary to neutralize 50.0 mL of a 16.0 M solution of sulfuric acid. The concentration of the NaOH is 2.50 M.

**Solution:**

2NaOH + H_{2}SO_{4}---> Na_{2}SO_{4}+ 2H_{2}OCalculate moles of H

_{2}SO_{4}by using n = C x V:n = 16.0 mol/L x 50 mL = 800 millimolesNow look at the equation. It says that for every mole of H

_{2}SO_{4}you need twice that many moles of NaOH to neutralize all the H_{2}SO_{4}.So moles of NaOH = 800 x 2 = 1600 millimoles of NaOH needed.

Now that you have moles and a concentration you can find volume:

V = moles / concentrationV = 1600 millimoles / 2.50 mmoles/mL = 640 mL

Note: I copied this answer from Yahoo Answers and the original writer put the molarity as 2.50 moles/L. If you did that on an answer to a test, you might get a deduction for using the wrong units on the molarity. Moles does not cancel with millimoles, although the numerical answer is correct.

**Problem #4:** What is the citric acid concentration in a soda if it requires 32.27 mL of 0.0148 M NaOH to titrate 25.00 mL of soda?

**Solution:**

Citric acid has three acidic hydrogens, so I will use H_{3}Cit for the formula.H

_{3}Cit + 3NaOH ---> Na_{3}Cit + 3H_{2}OThe key is the 1 : 3 molar ratio between H

_{3}Cit and NaOHmoles NaOH ---> (0.0148 mol/L) (0.03227 L) = 0.000477596 mol

1 is to 3 as x is to 0.000477596 mol

x = 0.0001592 mol (of H

_{3}Cit)0.0001592 mol / 0.0250 L = 0.00637 M (to three sig figs)

**Problem #5:** Consider the unbalanced molecular equation in which citric acid, H_{3}C_{6}H_{5}O_{7}, is reacted with sodium hydroxide:

H_{3}C_{6}H_{5}O_{7}(aq) + NaOH(aq) ---> Na_{3}C_{6}H_{5}O_{7}(aq) + H_{2}O(ℓ)

If 62.7 mL of 1.20 M NaOH(aq) is titrated with 32.0 mL of H_{3}C_{6}H_{5}O_{7}(aq) what the molarity of H_{3}C_{6}H_{5}O_{7}(aq)?

**Solution:**

1) First, let us balance the equation:

H_{3}C_{6}H_{5}O_{7}(aq) + 3NaOH(aq) ---> Na_{3}C_{6}H_{5}O_{7}(aq) + 3H_{2}O(ℓ)

2) The key point is the 1:3 stoichiometric ratio between the citric acid and the NaOH.

moles NaOH ---> (1.20 mol/L) (0.0627 L) = 0.07524 mol3) Calculate the molarity of the citric acid:You need three moles of NaOH to neutralize every one mole of citric acid.

moles citric acid ---> 0.07524 mol / 3 = 0.02508 mol

0.02508 mol / 0.0320 L = 0.78375 MThree sig figs gives 0.784 M for the final answer.

**Problem #6:** 51.00 ml of phosphoric acid solution (H_{3}PO_{4}) reacts with 13.90 grams of barium hydroxide, Ba(OH)_{2}, according to the following balanced equation. What is the molarity of the phosphoric acid?

**Solution:**

2H_{3}PO_{4}+ 3Ba(OH)_{2}---> Ba_{3}(PO_{4})_{2}+ 6H_{2}Omoles Ba(OH)

_{2}: 13.90 g / 171.344 g/mol = 0.08112335 mol3 moles Ba(OH)

_{2}react with 2 moles H_{3}PO_{4}0.08112335 mol Ba(OH)

_{2}react with x mol H_{3}PO_{4}x = 0.0540822 mol

molarity of phosphoric acid: 0.0540822 mol/0.05100 L = 1.06 M

**Problem #7:** Calculate the molar concentration of phosphoric acid if 90.0 mL are required to neutralize 200. mL of 0.200 M calcium hydroxide solution.

**Solution:**

1) First, let us balance the equation:

2H_{3}PO_{4}(aq) + 3Ca(OH)_{2}(aq) ---> Ca_{3}(PO_{4})_{2}(s) + 6H_{2}O(ℓ)Notice that solid calcium phosphate is formed. This is immaterial to the following calculation.

2) There is a 2:3 molar ratio between H_{3}PO_{4} and Ca(OH)_{2}.

moles Ca(OH)_{2}---> (0.2 mol/L) (0.2 L) = 0.04 mol2 is to 3 as x is to 0.04

x = 0.02667 mol of H

_{3}PO_{4}

3) Calculate the molarity of the H_{3}PO_{4}:

0.02667 mol / 0.09 L = 0.296296 MRounded off to three sig figs, 0.296 M is the H

_{3}PO_{4}concentration.

**Problem #8:** What is the concentration of a Ca(OH)_{2} solution if 10.0 ml of 0.600 M H_{3}PO_{4} solution is required to completely neutralize 12.5 ml of the Ca(OH)_{2} solution?

**Solution:**

3Ca(OH)_{2}+ 2H_{3}PO_{4}---> Ca_{3}(PO_{4})_{2}+ 6H_{2}OThe key is to see the 3 : 2 molar ratio between Ca(OH)

_{2}and H_{3}PO_{4}.moles H

_{3}PO_{4}---> (0.600 mol/L) (0.0100 L) = 0.00600 mol3 is to 2 as x is to 0.00600 mol

x = 0.00900 mol (of Ca(OH)

_{2}required)0.00900 mol / 0.0125 L = 0.720 M

**Problem #9:** 60.0 mL of 0.10 M CaOH_{2} solution is mixed with 50.0 mL of 0.10 M HNO_{3}. What is the volume of 0.10 M H_{2}SO_{4} needed to neutralize the mixture?

**Solution #1:**

moles Ca(OH)_{2}---> (0.10 mol/L) (0.060 mL) = 0.0060 molmoles HNO

_{3}---> (0.10 mol/L) (0.050 mL) = 0.0050 molCa(OH)

_{2}+ 2HNO_{3}---> Ca(NO_{3})_{2}+ 2H_{2}OTwo HNO

_{3}are required to neutralize one Ca(OH)_{2}, so only 0.0025 mol of Ca(OH)_{2}will be neutralized.0.0060 mol minus 0.0025 mol = 0.0035 mol of Ca(OH)

_{2}remainingCa(OH)

_{2}and H_{2}SO4 react in a 1 to 1 molar ratio, so 0.0035 mol of H_{2}SO_{4}is required.volume H

_{2}SO_{4}---> 0.0035 mol / 0.10 mol/L = 0.035 L = 35.0 mL

**Solution #2:**

Write a balanced equation for the first step:Ca(OH)

_{2}+ 2HNO_{3}---> Ca(NO_{3})_{2}+ 2H_{2}OYou need 2 moles of HNO

_{3}for every mole of Ca(OH)_{2}. How much was provided?n = (60 mL) (0.10 M) = (0.060 L) (0.10 mol/L) = 0.006 mol of Ca(OH)_{2}

n = (50 mL) (0.10 M) = (0.050 L) (0.10 mol/L) = 0.005 mol of HNO_{3}If we react these fully, 0.005 moles of HNO

_{3}will react with 0.0025 moles of Ca(OH)_{2}. This leaves 0.006 - 0.0025 = 0.0035 moles of Ca(OH)_{2}. Note that we are left with aqueous Ca(OH)_{2}and those extra hydroxide ions will make the solution basic, so it does make sense that we need to add more acid to neutralize it. Now we write an equation for the next reaction:Ca(OH)

_{2}+ H_{2}SO_{4}---> CaSO_{4}+ 2H_{2}OWe need one mole of H

_{2}SO_{4}for every one mole of Ca(OH)_{2}. That makes it easy. We have 0.0035 moles of Ca(OH)_{2}left over, so we need to add 0.0035 moles of H_{2}SO_{4}.To get volume, we need to divide the number of moles by the molarity:

V = n / M = (0.0035 mol) / (0.10 M) = 0.035 L = 35 mL of H_{2}SO_{4}

**Problem #10:** 50.0 mL of 2% (w/w) solution of NaOH is neutralized completely with 0.50 M H_{2}SO_{4}. Calculate (a) molality, (b) molarity of the NaOH solution given the density is 1.08 g/mL, and (c) the volume of 0.50 M H_{2}SO_{4}

**Solution:**

2% (w/w) means 2 g of NaOH are present in every 100 g of solution.

Since we have 50 mL of solution, we need to know its mass:

50.0 mL times 1.08 g/mL = 54.0 g

grams of NaOH in the 54 g of solution:

2 is to 100 as x is to 54x = 1.08 g

moles of NaOH ---> 1.08 g / 40.0 g/mol = 0.027 mol

molarity ---> 0.027 mol / 0.050 L = 0.54 M

For molality, we need the mass of water:

54.0 g minus 1.08 g = 52.92 g

molality ---> 0.027 mol / 0.05292 kg = 0.51

For part (c):

H_{2}SO_{4}+ 2NaOH ---> Na_{2}SO_{4}+ 2H_{2}OWe know that 2 NaOH are required for every one H

_{2}SO_{4}, so do this:0.027 mol / 2 = 0.0135 mol (this is how much H

_{2}SO_{4}is neutralized by our 0.027 mol of NaOH)volume = 0.0135 mol / 0.50 mol/L = 0.027 L = 27 mL

**Problem #11:** 2.00 liters of NH_{3} at 30.0 °C and 0.200 atm is neutralized by 134 ml of H_{2}SO_{4} solution. Calculate the molarity of the acid solution.

**Solution:**

1) We need to get the moles of NH_{3}:

PV = nRT(0.200 atm) (2.00 L) = (n) (0.08206 L atm / mol K) (303 K)

n = 0.016087 mol

2) Now, the chemical equation for the neutralization:

2NH_{3}+ H_{2}SO_{4}---> (NH_{4})_{2}SO_{4}The key is that 2 NH

_{3}are required for every one H_{2}SO_{4}.0.016087 mol / 2 = 0.0080435 mol <--- that's how many moles of H

_{2}SO_{4}reacted

3) The molarity:

0.0080435 mol / 0.134 L = 0.0600 M

**Problem #12:** 25.0 mL NaHCO_{3} containing 4.20 g/L required 12.5 mL of HCl to completely react with it. Calculate the molarity of the HCl solution.

**Solution:**

1) We need to know how many moles of NaHCO_{3} are present. To do that, we first get grams:

4.2 g is to 1000 mL as x is to 25.0 mLx = 0.105 g

2) Then, moles:

0.105 g / 84.0059 g/mol = 0.0012499 mol

3) From the balanced equation:

NaHCO_{3}+ HCl = NaCl + H_{2}O + CO_{2}we see that the molar ratio between NaHCO

_{3}and HCl is one to one. This means that 0.0012499 mol of HCl reacted.

4) The molarity of the HCl:

0.0012499 mol / 0.0125 L = 0.100 M (to three sig figs)

**Problem #13:** 25.0 mL of a solution of an acid H_{x}A containing 0.10 mol of the acid in each 1000 mL of solution reacts with 75.0 mL of a solution of 0.10 M NaOH. What is the value of x?

**Solution:**

moles of NaOH ---> (0.10 mol/L) (0.0750 L) = 0.00750 molmoles of H

^{+}(from H_{x}A) must equal moles of OH^{-}moles H

_{x}A ---> (0.10 mol/L) (0.0250 L) = 0.0025 molx = 0.00750 / 0.00250 = 3

Another problem of this type: In the reaction between an acid H_{y}A and 0.100 mol dm^{-3} NaOH solution, 25.0 cm^{3} of a solution of 0.100 mol dm^{-3} H_{y}A reacts with 50.0 cm^{3} of the 0.100 mol dm^{-3} NaOH. What is the value of y? **Answer: y = 2**

**Problem #14:** (a) If you titrate a 5.00 ml sample of an unknown monoprotic acid, HA, with 27.0 mL of a 0.160 M sodium hydroxide solution. Compute the molarity of the weak acid to the appropriate number of significant figures. Show all work

(b) If the unknown acid had been H_{2}SO_{4} with the same concentration as HA (calculated above), how much NaOH would have been required for the titration?

**Solution:**

For (a), this is the reaction:HA + NaOH ---> NaA + H_{2}OThe key is that there is a 1:1 molar ratio between HA and NaOH.

moles NaOH ---> (0.160 mol/L) (0.02700 L) = 0.00432 mol

From the 1:1 ratio, we know that 0.00432 mol of HA got neutralized.

molarity of HA ---> 0.00432 mol / 0.00500 L = 0.864 M

For (b), write the chemical equation:

H_{2}SO_{4}+ 2NaOH ---> Na_{2}SO_{4}+ 2H_{2}ONotice that, for every ONE H

_{2}SO_{4}neutralized, TWO NaOH are required.The same molarity of H

_{2}SO_{4}(the 0.864 value) would required TWICE as much NaOH as compared to neutralizing 0.864 M HA.54.0 mL is the answer.

**Problem #15:** You dilute 20.0 ml L of a stock solution of H_{2}SO_{4} to 1.00 L. You then make duplicate titrations of 20.0 mL each of the diluted H_{2}SO_{4} solution with your 0.202 M NaOH solution to the end point. Titration 1 requires 39.05 mL; titration 2 requires 39.09 mL. Calculate the concentration of the stock H_{2}SO_{4} solution.

**Solution:**

1) Use the average of the two volumes of NaOH:

(39.05 mL + 39.09 mL) / 2 = 39.07 mL

2) Moles of NaOH used:

(0.202 mol/L) (0.03907 L) = 0.00789214 mol

3) The balaced chemical equation is:

2NaOH + H_{2}SO_{4}---> Na_{2}SO_{4}+ 2H_{2}OTwo NaOH are used for every one H

_{2}SO_{4}neutralized.

4) Moles of H_{2}SO_{4} neutralized:

0.00789214 mol / 2 = 0.00394607 mol

5) The amount of H_{2}SO_{4} just above was in 20.0 mL and this volume came from the 1.00 L. How many moles of H_{2}SO_{4} are present in the 1.00 L?

0.00394607 mol is to 0.0200 L as x is to 1.00 Lx = 0.1973035 mol

6) All the above moles of H_{2}SO_{4} was originally contained in the 20.0 mL that was then diluted to 1.00 L. What is the molarity of the original 20.0 mL of stock solution?

0.1973035 mol / 0.0200 L = 9.865175 MTo three sig figs, 9.86 M

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