Question

32. What substances are present in an aquesous solution that results from mixing 25.0 ml of 0.20 M HCN with 20.0 mL of 0.25 M KOH? A) HCN, KOH and water B) HCN, KCN and water C) KCN, KOH, and water D) KCN and water 33. Calculate the pH of the solution resulting from the addition of 10.0 mL of 0.10 M NaOH to 50.0 mL of 0.10 M HCN (K. = 4.9 x 10-10) solution. A) 5.15 B) 8.71 C) 5.85 D) 9.91 E) 13.0

Answer 1

Question 32

mmol HCN = concentration $\times$ Volume = 0.20 M $\times$ 25.0 ml = 5.0 mmol

mmol KOH = concentration $\times$ Volume = 0.25 M $\times$ 20.0 ml = 5.0 mmol

Consider reaction, HCN (aq) + KOH (aq) $\rightarrow$ KCN (aq) + H2O (l)

According to reaction, 1 mol HCN reacts with 1 mol KOH and produces 1 mol KCN and 1 mol H2O.

Therefore, 5.0 mmol HCN will react with 5.0 mmol KOH and produce 5.0 mmol KCN and 5.0 mmol  H2O.

In this case, all HCN is consumed by added KOH.

After completion of reaction, solution contain only KCN and Water.

ANSWER : D) KCN and water.

Question 33

mmol HCN = concentration $\times$ Volume = 0.10 M $\times$ 50.0 ml =5.0 mmol

mmol KOH = concentration $\times$ Volume = 0.10 M $\times$ 10.0 ml =1.0 mmol

Consider reaction, HCN (aq) + KOH (aq) $\rightarrow$ KCN (aq) + H2O (l)

According to reaction, 1 mol HCN reacts with 1 mol KOH and produces 1 mol KCN and 1 mol H2O.

From above calculations, it is clear that HCN is excess reactant and KOH is limiting reactant.

Excess mmol HCN = 5.0 mmol - 1.0 mmol = 4.0 mmol

mmol KCN produced = mmol of KOH added = 1.0 mmol

Volume of solution after addition of KOH = 50.0 + 10.0 = 60.0 ml

Let's calculate [ HCN ] and [ KCN ]

[ HCN ] = 4.0 mmol / 60.0 ml = 0.0667 M

[ KCN ] = 1.0 mmol / 60.0 ml = 0.0167 M

After reaction solution contain weak acid HCN and its salt KCN . This solution acts as a buffer solution.

We can calculate pH of buffer solution by using Henderson's equation, pH = pKa + log [ Salt ]/ [ Acid ]

pH = - log Ka + log [ KCN ] / [ HCN ]

pH = - log ( 4.9 $\times$ 10 -10 ) + log 0.0167 / 0.0667

pH = 9.31 + log 0.0167 / 0.0667

pH = 9.31 - 0.601

pH = 8.71

ANSWER : B) 8.71