The reaction of the weak acid HCN with the strong base KOH is: HC N(aq) + KO H {

Question

The reaction of the weak acid HCN with the strong base KOH is: HC N(aq) + KO H {aq) rightarrow HOH (l) + KCN(aq) To compute the pH of the resulting solution if 73mL of 0.70M HCN is mixed with 16mL of 0.30 KOH we need to start with the stoichiometry. Lets do just the stoich in 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? These are the initial concentration in an ICE Table!

Explanation / Answer

moles of acid = 0.7 * 73 / 1000 = 0.0511 mol

Mole sof base = 0.30 * 16 / 1000 = 0.0048 mol

Limiting reagent = Base, KOH

Final moles of Acid = 0.0511 - 0.0048 = 0.0463 mol

Final concentration of acid = 0.0463 / (73 + 16) = 0.000520 M

Concentration of salt formed (KCN) = 0.0048 / 89 = 0.0000539 M

pKa of acid, HCN = 9.2

Here, a buffer of HCN and KCN is formed.

pH fromual for buffer solutions,

pH = pKa + Log[salt]/[acid]

pH = 9.2 + Log(0.0000539 / 0.000520)

pH = 8,22

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