1)Calculate the pH of the solution after the addition of 8.58×10^{-2} mol of NaOH(s).

[H+] = (8580-8.58v)/(1000+v) * 10^(-2), where v is the volume of NaOH added

v = 149 mL

and:

[H+] = 0.06354

so:

pH = 1.197

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