Question

A metal plating solution contains 50.00 mg/L of copper. Determine the concentration, in moles/L, to which the hydroxide concentration must be raised to precipitate all but 1.3 mg/L of the copper using lime. The Ksp of copper hydroxide is 2.00 3 10219. Estimate the final pH (report your answer to two decimal places).

Answer 1

Answer:

$pH=6.51$

Explanation:

Given:

• The concentration of Cu initially = $50mgL^{-1} = \frac{50*10^{-3}}{63.5}=0.787*10^{-3}molL^{-1}$

• The concentration of Cu finally = $1.3mgL^{-1} = \frac{1.3*10^{-3}}{63.5}=0.0205*10^{-3}molL^{-1}$
• $K_{sp}=2.20*10^{-20}$ -This is the actual $K_{sp}$ of $Cu(OH)_2$

We know that ,

$Cu(OH)_2$⇒$Cu^{2+}+2OH^-$

Therefore ,

$K_{sp}=[Cu^{2+}][OH^-]^2$

$2.20*10^{-20}=0.0205*10^{-3}*[OH^-]^2\\1.0731*10^{-15}=[OH^-]^2\\3.276*10^{-8}=[OH^-]$

$pOH=-log[OH^-]\\\\pOH=-log(3.276*10^{-8})\\pOH=7.485\\pH=14-pOH\\pH=14-7.49\\pH=6.51$

Answer: $pH=6.51$