Question

**********PLEASE SHOW WORK *********
1) You have a solution of acetic acid that has a K a of 3.5 × 10 –8 . If the concentration of the acetic acid were 2.40 M, what would be the concentration of H + at equilibrium?
2) You have a solution that is 1.75 M HCN. If the K a is 9.9 × 10 –8 , calculate the pH of the solution.
3) How many grams of KOH are needed to neutralize 35.0 mL of 1.8 M HCl?
4) How many mL of 2.20 M KOH are needed to neutralize 20.0 mL of 1.40 M HNO 3 ?

Answer 1

Answer:

1) $[H^+]_{eq}=0.00029M$

2) $pH=3.38$

3) $m=3.53gKOH$

4) $V_{KOH}=12.7mL$

Explanation:

Hello,

1) Here, you should remember acetic acid dissociation is:

$CH_3COOH(aq)+H_2O(l)\rightleftharpoons CH_3COO^-(aq)+H_3O^+(aq)$

Thus, we propose the law of mass action for the aqueous species only:

$Ka=\frac{[CH_3COO^-][H_3O^+]}{[CH_3COOH]}$

Hence, in terms of the change $x$ due to reaction's extent we have:

$3.5x10^{-8}=\frac{x*x}{2.40-x}$

Solving for $x$ we obtain 0.00029 which equals the concentration of H⁺ or H₃O⁺ (they are the same) at equilibrium:

$[H^+]_{eq}=0.00029M$

2) As well as in the previous case, we write its equilibrium dissociation:

$HCN(aq)+H_2O(l)\rightleftharpoons CN^-(aq)+H_3O^+(aq)$

Thus, we propose the law of mass action for the aqueous species only:

$K=\frac{[CN^-][H_3O^+]}{[HCN]}$

Hence, in terms of the change $x$ due to reaction's extent we have:

$9.9x10^{-8}=\frac{x*x}{1.75-x}$

Solving for $x$ we obtain 0.000416 which equals the concentration of H⁺ or H₃O⁺ (they are the same) at equilibrium:

$[H^+]_{eq}=0.000416 M$

So we compute the pH:

$pH=-log([H^+]_{eq})=-log(0.000416)\\\\pH=3.38$

3) In this case, from the formula defining molarity we first compute the moles of KOH whose molar mass is 56.1 g/mol considering the volume in litres:

$M=\frac{n}{V}\\ \\n=1.8\frac{mol}{L} *35.0mL*\frac{1L}{1000mL} =0.063molKOH$

So we compute the grams by using its molar mass:

$m=0.063molKOH*\frac{56.1gKOH}{1molKOH} \\m=3.53gKOH$

4) In this case, for this neutralization as both of them have one equivalent (one hydrogen in the acid and one hydroxyl in the base), we solve the following formula for the volume of KOH as the moles become equal at the equivalence:

$M_{HNO_3}V_{HNO_3}=V_{KOH}\\\\V_{KOH}=\frac{M_{HNO_3}V_{HNO_3}}{M_{KOH}} =\frac{1.40M*20.0mL}{2.20M}\\ \\V_{KOH}=12.7mL$

Best regards.