Question

the reaction of the strong acid HCN with the strong base KOH is:

HCN(aq) + KOH(aq) --> HOH(l) + KCN(aq)

To compute the pH of the resulting solution if 63mL of 0.67M HCN is mixed with 26mL of 0.41M KOH. Let's do this in using stoich steps:

How many moles of acid?

How many moles of base?

What is the limiting reactant?

How many moles of the excess reactant after reaction?

What is the concentration of the excess reactant after reaction?

What is the concentration of the pH active product after reaction?

Answer 1

Given:

M(HCN) = 0.67 M

V(HCN) = 63 mL

M(KOH) = 0.41 M

V(KOH) = 26 mL

1)

mol(HCN) = M(HCN) * V(HCN)

mol(HCN) = 0.67 M * 63 mL = 42.21 mmol

Answer: 0.04221 mol of acid

2)

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.41 M * 26 mL = 10.66 mmol

Answer: 0.01066 mol of acid

3)

We have:

mol(HCN) = 42.21 mmol

mol(KOH) = 10.66 mmol

10.66 mmol of both will react

KOH is limiting

4)

remaining mol of HCN = 31.55 mmol

Answer: 0.03155 mol of HCN remains

5)

Total volume = 89.0 mL

[H+]= mol of acid remaining / volume

[H+] = 31.55 mmol/89.0 mL

= 0.354494 M

use:

pH = -log [H+]

= -log (0.3545)

= 0.450

Answer: 0.450