Compute the pH of the solution after reaction of 53 mL of 0.65M HCN with 34mL of 0.39M KOH

Question

HCN(aq)+KOH(aq)———-HOH(l)+KCN(aq)

compute the moles of reactant before reaction

1)moles of acid____moles

2)moles of base _____moles

moles of reactant and pH active after reaction

excess of reactant ____moles

moles of product_____moles

now convent to concentration units

reactant=_____ molar

product=_____molar

set up an ICE table

and use the equilibrium concentration from ICE table and make the appropriate approximations to fill in the table below then solve for x and pH K of HCN is 3.5 X 10^-4

now compute pH

H+=______ M

pH=______

fill in the blanks please

Answer 1

HCN + KOH ==> KCN + H2O ... BALANCED EQUATION

moles HCN before reaction = 53 ml x 1 L/1000 ml x 0.65 moles/L = 0.03445 moles HCN

moles KOH before reaction = 34 ml x 1 L/1000 ml x 0.39 moles/L = 0.01326 moles KOH

Since the HCN and KOH react in a 1:1 mole ratio (see balanced equation), all the KOH will be used up, and there will be excess HCN.

The amount of excess HCN = 0.03445 moles - 0.01326 moles = 0.02119 moles HCN after reaction

Moles of product (KCN) formed = 0.01326 moles KCN after reaction (1:1 mol ratio from balanced equation)

Converting these to concentration units, we have a total volume of 53 ml + 34 ml =87 mls = 0.087 L

[HCN] = 0.02119 moles / 0.087 L = 0.244 M

[KCN] = 0.01326 moles / 0.087 L = 0.152 M

THIS IS A BUFFER

From the Ka of 3.5x10^-4 for HCN we can find the pKa

pKa = -log Ka = -log 3.5x10^-4 = 3.46

I will use the Henderson Hasselbalch equation to find the pH

pH = pKa + log [KCN] / [HCN] = 3.46 + log (0.152/0.244)

pH = 3.46 -0.21

pH = 3.25

[H+] = 5.6x10^-4 M