A 25.0 ml sample of 0.150 m butanoic acid is titrated with a 0.150 m naoh solution. what is the ph after 26.0 ml of base is added? the ka of butanoic acid is 1.5 ⋅ 10-5. a 25.0 ml sample of 0.150 m butanoic acid is titrated with a 0.150 m naoh solution. what is the ph after 26.0 ml of base is added? the ka of butanoic acid is 1.5 10-5. 4.84 2.54 7.00 4.81 11.47

First,

Moles of HBu = Volume per liter * moles of HBu / liter

= 25ml/1000 * 0.15 = 0.00375 moles

moles of NaOH = volume per liter * moles of NaOH/liter

= 1 ml/1000 * 0.15 = 0.00015 moles

according to this equation:

HBu + NaOH → H2O + NaBu

when 1 mol of NaOH gives 1mol of HBu

So 0.00015 of NaOH will give 0.00015 mol of HBu

∴moles of HBu which remains = 0.00375 - 0.00015 = 0.0036 moles

∴moles of Bu - produced = 0.00015 moles

when the total volume = 0.025 + 0.026 = 0.051 L

[HBu] = 0.0036moles / 0.051 L = 0.071 moles

[Bu] = 0.00015 / 0.051L = 0.0029 moles

when Ka = [H+] [Bu] / [HBu]

1.5x10^-5 = [H+] (0.0029) / (0.071)

∴[H+] = 1x10^-6 / 0.076 = 1.5 x 10^-5

∴PH = - ㏒[H+]

= - ㏒ 1.5x10^-5

= 4.81