Question

Calculate the pH when 65.0 mL of 0.269 M or a certain monoprotic weak acid, HA, is mixed with 65.0 mL of 0.269 M sodium hydroxide solution at 25 degree C. For HA, the Ka is 1 .5x 10^-5.

Answer 1

NaOH reacts with HA in 1:1 mol ratio
Mol acid in 65mL of 0.269 M solution = 65/1000 * 0.269 = 0.017485 mol
Mol NaOH in 65mL of 0.269 M solution = 65/1000 * 0.269 = 0.017485 mol NaOH

When these react you will produce 0.017485 mol NaA and there will be 0.017485 - 0.017485 = 0 mol HA left unreacted
The final volume of solution = 65mL + 65mL = 130 mL = 0.13 L
[salt] = 0.017485 / 0.13 = 0.1345 M
[acid] = 0 / 0.13 = 0 M

pKa acid = -log (1.5*10^-5) = 4.824
Use Henderson - Hasselbalch equation:
pH = pKa + log ([salt]/[acid]
pH = 4.824 + log (0.1345 / 0)
The right hand side value is not valid hence
pH = pKa = 4.824

Answer 2

Reaction of weak acid with strong base is Number of moles of 0.269 M x65.0 mLx 1000 mL 0.0175 Number of moles of NaOH= 0.2 9 M 5.0 mL× 1000 mL 0.0175 Total volume 5.0 mL + 5.0 mL =130 mL 130 mLx 1000mL =0.13L 0.0175 0.13 -0.135 M Concentration ofNaK. moles/L 1.0 x10-14 ,-1.5×10 =6.7×10-10 0.135 0.135-X 0 0 -5-0.135-x ㄨㄨㄨ 0.135-x 10 6.7×10-10- (0.135-x~ 0.135 X2 =6.7 × 10-10 × 0.135 0.9 ×10 x=| OH- 1=0.95× 10- =-log OH . log(0.95x10-5 =5-0.22 = 4.98 pH-14-рон -14-4.98 -9.02