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Answer:
a) pH = 11.5
b) pH= 3
Explanation:
a) lets calculate the number of moles of each reactant
moles of HCl = 32/1000 * 0.32 = 0.01024 mole
moles of NaOH = 42/1000 * 0.32 = 0.01344 moles
1 moles of HCl reacts with 1 moles of NaOH , so 0.01024 mole of HCl should react with 0.01024 moles of NaOH , but there is some excess NaOH.
excess NaOH= 0.01344 -0.01024 = 0.0032 moles
[H+]=
pH= -log [3.125*10^-12) = 11.5
b) moles of NaOH = 0.00924
excess HCl present = 0.01024 - 0.00924 =.001
so excess [H+] = 0.001
pH= -log( 0.001) = 3