What is the pOH if a solution composed of 60mL of 0.0012M HCl and 200 mL of 0.0005M NaOH?

Question

Angel Velazquez

Biology

15 September, 23:15

What is the pOH if a solution composed of 60mL of 0.0012M HCl and 200 mL of 0.0005M NaOH?

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Answers (1)

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Karla Gaines · Answer 1

Moles = Molarity * Volume

For HCl, Moles = 60 x 10^-3 * 1.2 x 10^-3 = 7.2 x 10^-5 moles

For NaOH Moles = 200 x 10^-3 * 5 x 10^-4 = 10 x 10^-5 moles

NaOh + HCl - > NaCl + H2O

7.2 x 10^-5 moles HCl reacts with 7.2 x 10^-5 moles NaOH to form NaCl and H2O

After reaction, Moles of NaOH left = 10 x 10^-5 - 7.2 x 10 ^-5 = 2.8 x 10^-5 moles

Now volume = 260 mL = 200 + 60. So, [OH-] = 2.8 x 10^-5 / 260 x 10^-3 = 0.010769 x 10^-2 = 1.0769 x 10^-4

pOH = 3.679