Question

Calculate the pH of the resulting solution if 32.0 mL of 0.320 M HCl(aq) is added to 42.0 mL of 0.320 M NaOH(aq).

pH=

Calculate the pH of the resulting solution if 32.0 mL of 0.320 M HCl(aq) is added to22.0 mL of 0.420 M NaOH(aq).

pH=

Answer 1

1)

Moles of HCl = (0.320mol/1000ml) × 32.0ml = 0.01024mol

moles of NaOH = (0.320mol/1000ml) × 42ml = 0.01344mol

NaOH + HCl -------> NaCl + H2O

1:1 molar reaction

0.01024moles of HCl reacts with 0.01024moles of NaOH

remaining moles of NaOH = 0.01344mol - 0.01024mol = 0.0032mol

moles of OH^-

Total volume = 74ml

[OH^-] = (0.0032mol/74ml)×1000ml = 0.04324M

pOH = -log[OH^-]

pOH = - log(0.04324M)

pOH = 1.36

pH = 14 - pOH

pH = 14 - 1.36

pH = 12.64

2)

Moles of HCl = 0.01024mol

moles of NaOH = ( 0.420mol/1000ml) × 22ml = 0.00924mol

0.00924mol of NaOH react with 0.00924moles of HCl

remaining moles of HCl = 0.01024mol - 0.00924mol = 0.0010mol

moles of H⁺ = 0.0010mol

Total volume = 54ml

[H⁺] = ( 0.0010mol/54ml) ×1000ml = 0.01852M

pH = -log[H⁺]

pH = -log(0.01852M)

pH = 1.73