Molarity of the copper (I) hydroxide

Question

A student uses acetic acid to titrate a solution of copper(I) hydroxide to determine the concentration of the copper (I) hydroxide. If 30.0 mL of 0.250 M acetic acid is needed to neutralize 15.0 mL copper (I) hydroxide, what is the molarity of the copper(I) hydroxide?

Answer 1

This is a reaction between a strong base (OH-) and a weak acid (acetic acid).

The overall reaction is

Cu(OH) (aq) + CH₃COOH (aq)---> H₂O (l) + Cu(CH₃COO) (aq)

Make sure the equation is balanced. (This is balanced)

How much acid was used?

30 ml of 0.250 M

0.030 L * 0.250 M = 0.0075 moles of acid

Stoichiometric ratio between acid and base is 1:1

S0 there were 0.0075 moles of OH^- present at the start.

Which is also 0.0075 moles of CuOH (aq) by stoichiometry

These 0.0075 moles were in an original volume of 15.0 ml (0.015 L)

So the molarity of the CuOH is 0.0075 / 0.015 = 0.500 M

Let me know if you have any questions!