Problems 1 - 10

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**Problem #1:** Calculate the pH of the solution that results when 40.0 mL of 0.100 M NH_{3} is:

a) Diluted with 20.0 mL of distilled water.

b) Mixed with 20.0 mL of 0.200 M HCl solution.

c) Mixed with 20.0 mL of 0.200 M NH_{4}Cl

**Solution to part (a):**

1) Use the dilution equation:

M_{1}V_{1}= M_{2}V_{2}(0.100 mol/L) (0.040 L) = (x) (0.060 L)

x = 0.0667 M

2) Use the K_{b} for ammonia to determine pH:

K_{b}= ([NH_{4}^{+}] [OH¯]) / [NH_{3}]1.77 x 10¯

^{5}= (x) (x) / (0.0667 - x)Remember to ignore the x in 0.0667 - x.

x = 0.001086278 M (a few guard digits)

pOH = - log 0.001086278 = 2.964

pH = 11.036

**Solution to part (b) and (c):** Someday!

**Problem #2:** How many grams of ammonia are needed to make 1.25 L solution with a pH of 11.68?

**Solution:**

1) Use the pH to get the hydroxide ion concentration:

pH = 11.68pOH = 14.00 - 11.68 = 2.32

[OH¯] = 10¯

^{pOH}= 10¯^{2.32}[OH¯] = 4.7863 x 10¯

^{3}

2) Use the K_{b} expression to get the [NH_{3}]:

K_{b}= ([NH_{4}^{+}] [OH¯]) / [NH_{3}]1.77 x 10¯

^{5}= (4.7863 x 10¯^{3}) (4.7863 x 10¯^{3}) / (x)x = 1.294275 M (guard digits!)

3) Grams needed for 1.25 L

MV = g/molar mass(1.294275 mol/L) (1.25 L) = x/17.031 g/mol

x = 27.6 g

**Problem #3:** Calculate the pH of a formic acid solution that contains 1.45% formic acid by mass. (Assume a density of 1.01 g/mL for the solution.)

**Solution:**

1) Calculate volume of 100.0 g of solution:

100.0 g divided by 1.01 g/mL = 99.01 mL

2) Calculate molarity of 1.45% solution of HCOOH:

1.45 g divided by 46.025 g/mol = 0.0315 mol0.0315 mol / 0.09901 L = 0.318 M

3) Calculate [H^{+}] in 0.318 M HCOOH solution:

1.8 x 10^{-4}= [(x) (x)] / 0.318x = 7.5657 x 10

^{-3}M (kept a couple guard digits)

4) Calculate pH:

-log 7.5657 x 10^{-3}= 2.12

**Problem #4:** A buffer solution contains 0.384 M KHCO_{3} and 0.239 M Na_{2}CO_{3}.
If 0.0464 moles of potassium hydroxide are added to 225.0 mL of this buffer, what is the pH of the resulting solution ? (Assume that the volume does not change upon adding potassium hydroxide.)

**Solution:**

1) Calculate moles of bicarbonate and carbonate:

HCO_{3}¯: (0.384 mol/L) (0.2250 L) = 0.0864 mol

CO_{3}^{2}¯: (0.239 mol/L) (0.2250 L) = 0.053775 mol

2) The hydroxide reacts with the acid (the bicarbonate):

HCO_{3}¯ decreases: 0.0864 - 0.0464 = 0.0400 mol

CO_{3}^{2}¯ increases: 0.053775 + 0.0464 = 0.100175 mol

3) Need pK_{a} of bicarbonate:

K_{a}of HCO_{3}¯ is the same as the K_{a2}of H_{2}CO_{3}.K

_{a}of HCO_{3}¯ = 4.7 x 10¯^{11}pK

_{a}= 10.252

4) Use Henderson-Hasselbalch Equation:

pH = 10.328 + log (0.100 / 0.04)pH = 10.328 + 0.398 = 10.726

**Problem #5:** A buffer solution contains 0.348 M ammonium chloride and 0.339 M ammonia. If 0.0248 moles of hydrochloric acid are added to 125.0 mL of this buffer, what is the pH of the resulting solution? (Assume that the volume does not change upon adding hydrochloric acid.)

**Solution:**

A solution on video is provided for this problem.

**Problem #6:** How many mL of 4.50 M sodium hydroxide must be added to 250.0 mL of a 0.200 M acetic acid solution to make a buffer with pH = 5.000?

**Solution:**

1) Use H-H Equation to determine required ratio of acetate to acid in solution:

5.000 = 4.752 + log [base] /[acid]log [base] /[acid] = 0.248

[base] /[acid] = 1.77

2) Determine molar amount of base required to get pH = 5.000 (for convenience, I'm going to use 1.00 L. I'll go to 250 mL at the end of this step):

1.77x + x = 0.200x = 0.0722 mol (this is acetic acid needed in the solution)

0.200 - 0.0722 = 0.1278 mol of base required

0.1278 /4 = 0.03195 mol of acetate required in 250 mL

3) Determine volume of NaOH solution required:

4.50 mol/L = 0.03195 mol / xx = 0.0071 L = 7.1 mL

4) Check everything:

[acetic acid] = 0.01805 mol / 0.2571 L = 0.070206 M[acetate] = 0.03195 mol / 0.2571 L = 0.12427 M

pH = 4.752 + log (0.12427 / 0.070206)

pH = 4.752 + log 1.77

pH = 4.752 + 0.248 = 5.000

**Problem #7:** If in a solution 10[H_{3}O^{+}] = [OH¯] then the pH of the solution is:

**Solution:**

1) Let:

[H_{3}O^{+}] = x

[OH¯] = y

2) Therefore, two equations in two unknowns:

10x = yand

xy = 1.00 x 10¯

^{14}

3) Rearrange the second equation:

y = (1.00 x 10¯^{14}) / x

4) Substitute into the first equation:

10x = (1.00 x 10¯^{14}) / x

5) Rearrange and solve:

10x^{2}= 1.00 x 10¯^{14}x

^{2}= 1.00 x 10¯^{15}x = 3.16 x 10¯

^{8}M

6) Compute the pH:

pH = -log 3.16 x 10¯^{8}= 7.5

**Problem #8:** The solubility of CO_{2}(g) in pure water is 0.0037 mol/L. Assuming that dissolved CO_{2} is in the form of H_{2}CO_{3}(aq), what is the pH of a 0.0037 M solution of dissolved CO_{2}? K_{a1} for H_{2}CO_{3} = 4.3 x 10^{-7}

**Solution:**

1) The relevant chemical equation is:

H_{2}CO_{3}⇌ H^{+}+ HCO_{3}¯

2) Substituting into the K_{a} expression, we find:

4.3 x 10^{-7}= [(x) (x)] / 0.0037x = 3.989 x 10

^{-5}M

3) Determine the pH:

pH = - log [H^{+}] = -log 3.989 x 10^{-5}pH = 4.40 (to two sig figs)

**Problem #9:** A 0.484 g sample of impure NH_{4}Br is treated with 25.00 mL of 0.2050 M NaOH and heated to drive off the NH_{3}. The unreacted NaOH in the reaction mixture after heating required 9.95 mL of 0.0525 M H_{2}C_{2}O_{4} to neutralize. How many grams of NH_{4}Br were in the original sample?

**Solution:**

1) Determine moles of NaOH added to amonium bromide solution:

(0.2050 mol/L) (0.02500 L) = 5.125 x 10¯^{3}molKeep in mind that only some of this NaOH reacted.

2) Determine moles of oxalic acid that reacted with excess NaOH:

(0.0525 mol/L) (0.00995 L) = 5.22375 x 10¯^{4}mol

3) Determine moles of unreacted NaOH:

This reaction:Hdemonstrates a 1:2 molar ratio between oxalic acid and NaOH._{2}C_{2}O_{4}(aq) + 2NaOH(aq) ---> Na_{2}C_{2}O_{4}+ 2H_{2}O(l)Therefore, we multiply the oxalic acid amount by two:

5.22375 x 10¯to determine the amount of NaOH that did not react with the ammonium bromide.^{4}mol x 2 = 1.04475 x 10¯^{3}mol

4) Determine moles of NaOH that did react with NH_{4}Br:

5.125 x 10¯^{3}mol minus 1.04475 x 10¯^{3}mol = 4.08025 x 10¯^{3}mol

5) Determine grams of NH_{4}Br:

4.08025 x 10¯^{3}mol times 97.943 g/mol = 0.399632 gTo three sig figs, this is 0.400 g

**Example #10:** If 0.50 moles Ca(OH)_{2} is slurried in 0.50 L deionized water and treated with 0.50 moles of CO_{2} gas in a closed system, the liquid phase of this system will have a pH closest to what value?

**Solution:**

After the Ca(OH)_{2}and the CO_{2}react, we are left with some calcium carbonate, an insoluble substance. However, from the K_{sp}of CaCO_{3}, we can calculate the approximate molarity of carbonate in the aqueous phase. Please see here for a discussion:I will use the 5.5 x 10

^{-5}M from the above link.Carbonate is the salt of a weak acid and so it hydrolyzes in solution:

CO

_{3}^{2-}+ H_{2}O ⇌ HCO_{3}^{-}+ OH^{-}To describe that system, we require the K

_{b1}of carbonate, which we get from the K_{a2}of carbonic acid, which is 4.7 x 10^{-11}, from here.So the K

_{b1}of carbonate is 2.13 x 10^{-4}(from K_{a}K_{b}= K_{w})We can now calculate the [OH

^{-}] in our calcium carbonate solution:[OH

^{-}] = SQRT[(2.13 x 10^{-4}) (5.5 x 10^{-5})] = 0.000108 MThe pOH is just under 4, which makes the pH be just over 10.

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