25.0 mL of a HCl solution was titrated with 0.123 mol L-1 NaOH. It
was found that 23.5 mL of NaOH was required to reach the reaction.
Dtermine pH of the solution.
Solution:
HCl + NaOH NaCl + H2O
Moles of HCl = moles of NaOH
25.0 x 10-3 x x = 0.123 x 23.5 x 10-3
x = 0.116
pH = -log10(0.116) = 0.937
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