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 adde
d? 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
