Problems #11 - 20

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**Problem #11:** 60.0 mL of 0.10 M Ca(OH)_{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 #12:** 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 #13:** 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 #14:** 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 #15:** 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

**Problem #16:** 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?

**Solution:**

moles of NaOH ---> (0.100 mol/L) (0.0500 L) = 0.00500 molmoles of H

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

_{y}A ---> (0.10 mol/L) (0.0250 L) = 0.0025 moly = 0.00500 / 0.00250 = 2

**Problem #17:** (a) You titrate a 5.00 mL sample of a generic monoprotic acid, HA, with 27.0 mL of a 0.160 M sodium hydroxide solution. Compute the molarity of the acid to the appropriate number of significant figures.

(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 #18:** 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

**Problem #19:** 12.0 mL of 0.100 M NaOH is mixed with 48.0 mL of Ba(OH)_{2} of unknown concentration. The solution is completely neutralized by 47.7 mL of 0.0500 M H_{2}SO_{4}. What is the concentration of the original Ba(OH)_{2} solution?

**Solution:**

1) We will first ignore the presence of the Ba(OH)_{2} solution and determine how much H_{2}SO_{4} was neutralized by the NaOH.

moles NaOH ---> (0.1 mol/L) (0.012 L) = 0.0012 molTwo moles of NaOH are required for every one mole of H

_{2}SO_{4}.moles H

_{2}SO_{4}neutralized ---> 0.0012 mol / 2 = 0.0006 molvolume H

_{2}SO_{4}used ---> 0.0006 mol / 0.05 mol/L = 0.012 L

2) The rest of the volume of H_{2}SO_{4} was used to neutralize the Ba(OH)_{2} solution.

47.7 - 12.0 = 35.7 mL

3) We now require the moles of Ba(OH)_{2}. We get this by using the remaining H_{2}SO_{4}.

moles H_{2}SO_{4}---> (0.05 mol/L) (0.0357 L) = 0.001785 mol

4) The H_{2}SO_{4} and Ba(OH)_{2} react in a 1:1 molar ratio. Therefore:

moles Ba(OH)_{2}= 0.001785 molmolarity of Ba(OH)

_{2}---> 0.001785 mol / 0.048 L = 0.0372 M

**Problem #20:** A 20.0 mL solution of triprotic acid is titrated to neutralization by 42.0 mL of 0.0483 M KOH solution. What is the molarity of the acid solution?

**Solution:**

1) Call the triprotic acid H_{3}A. Its reaction with KOH is this:

H_{3}A + 3KOH ---> K_{3}A + 3H_{2}OWe see three moles of KOH are required for each mole of H

_{3}A that reacts.

2) The relationship between amounts of acid and base that are required for neutralization is codified here:

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

3) Insert values in place and solve:

(x) (20.0 mL) (0.0483 M) (42.0 ml) ––––––––––– = –––––––––––––––– 1 3 x = 0.0338 M

**Problem #21:** A 20.0 mL solution of triprotic acid is titrated to neutralization by 42.0 mL of 0.0483 M Al(OH)_{3} solution. What is the molarity of the acid solution?

**Solution:**

1) Call the triprotic acid H_{3}A. Its reaction with Al(OH)_{3} is this:

H_{3}A + Al(OH)_{3}---> AlA + 3H_{2}OWe see one mole of Al(OH)

_{3}is required for each mole of H_{3}A that reacts.

2) The relationship between amounts of acid and base that are required for neutralization is codified here:

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

3) Insert values in place and solve:

(x) (20.0 mL) (0.0483 M) (42.0 ml) ––––––––––– = ––––––––––––––––– 1 1 x = 0.101 M (to three sig figs)

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