Question

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The pH of a solution prepared by mixing 50.0 mL of 0.125 M KOH and 50.0 mL of 0.125 M HCl is __________. Answer 7.00

A 25.0 mL sample of an acetic acid solution is titrated with a 0.175 M NaOH solution. The equivalence point is reached when 37.5 mL of the base is added. The concentration of acetic acid is __________ M.

Answer 0.263

Answer 1

1)

millimoles of KOH = 50 x 0.125 = 6.25

millimoles of HCl = 50 x 0.125 = 6.25

at equivalence point

millimoles of acid = millimoles of base

pH = 7.00

2)

millimoles of NaOH = 0.175 x 37.5 = 6.5625

volume of acetic acid = 25.0 mL

at equivalence point

millimoles of NaOH = millimoles of acetic acid

6.5625 = 25 x M

M = 0.263

concentration of acetic acid = 0.263 M