Question

Calculate the pH of the resulting solution if 16.0 mL of 0.160 M HCl(aq) is added to (a) 21.0 mL of 0.160 M NaOH(aq). (b) 26.0 mL of 0.210 M NaOH(aq).

Answer 1

number of moles of acid = 16 * 0.16 = 2.56 m moles

number of moles of base = 21 * 0.16 = 3.36 m moles

1 mole of HCl requires 1 mole of NaOH

2.56 mole of HCl requires 2.56 moles of NaOH.

remaining number of moles of NaOH = 3.36 - 2.56 = 0.8 m moles

concentration of NaOH = (0.8 *10^-3) * 1000 / (16+21) = 0.02162 M

pOH = -log(NaOH)

pOH = -log ( 0.02162)

pOH = 1.6651

pH = 14 - 1.6651 = 12.3349

number of moles or NaOH = 26 * 0.21 = 5.46 m moles

reamaing moles of NaOH = 5.46 - 2.56 = 2.9 m moles

Concentration of NaOH = (2.9 * 10^-3 * 1000) / (16+26)

concentration of NaOH = 0.06904

pOH = -log(0.06904)

pOH = 1.1609

pH = 14 - 1.1609

pH = 12.8391