Question

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

Answer 1

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