In: Chemistry
11. Calculate the pH of a solution prepared by mixing
a) 48.0 mL of 0.212 M NaOH and 66.0 mL of 0.187 M HCl
b) 69.0 mL of 0.345 M KOH and 125 mL of 0.400 M HC3H5O2
c) 350 mL of 0.130 M NH3 and 400 mL of 0.100 M HCl
11)
a)
millimoles of NaOH = 48 x 0.212 = 10.176
millimoles of HCl = 66 x 0.187 = 12.342
here millimoles of acid > millimoles of base
[H+] = 12.342 - 10.176 / 48 + 66 = 0.019 M
pH = -log [H+] = -log (0.019)
pH = 1.72
b) 69.0 mL of 0.345 M KOH and 125 mL of 0.400 M HC3H5O2
millimoles of KOH = 69 x 0.345 = 23.805
millimoles of CH3COOH = 125 x 0.4 = 50
CH3COOH + KOH --------------------> CH3COOK + H2O
50 23.805 0 0
26.195 0 23.805
pH = pKa + log [salt / acid]
= 4.74 + lof [23.805 / 29.195]
pH = 4.70
c) 350 mL of 0.130 M NH3 and 400 mL of 0.100 M HCl
millimoles of NH3 = 350 x 0.130 = 45.5
millimoles of HCl = 400 x 0.1 = 40
NH3 + HCl -----------------> NH4Cl
45.5 40 0
5.5 0 40
pOH = pKb + log [salt / base]
= 4.74 + log [40 / 5.5]
pOH = 5.60
pH = 8.40