Calculate the pH of the resulting solution if 32.0 mL of 0.320 M HCl(aq) is added to (a) 42.0 mL of 0.320 M NaOH(aq). (b) 22.0 mL of 0.420 M NaOH(aq).

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

screening

explanation:

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