If 100. mL of 0.100 M NaOH is added to 50. mL of 0.10 M HCl,...

Question

Question

If 100. mL of 0.100 M NaOH is added to 50. mL of 0.10 M HCl, what will be the pH at 25∘C?

Round your answer to two decimal places.

Answer 1

Answer #1

HCl + NaOH ---------> NaCl + H2O

moles of HCl = 0.10 x 50 /1000 = 0.005

moles of NaOH = 0.1 x 100/1000 = 0.01

moles of NaOH left = 0.01 - 0.005 = 0.005

[NaOH] = 0.005 / 0.15 = 0.033 M

NaOH is strong base so [NaOH] = [OH^-] = 0.033 M

pOH = - log [OH^-]

pOH = - log [0.033]

pOH = 1.48

pH = 14 - 1.48

pH = 12.52

Answer 2

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