Answer:
The new concentration of hydroxide ions is \(1\times 10^{-10} M\).
Explanation:
Concentration of \(H^+\) ions = \([H^+]=1\times 10^{-7} M\)
When concentration of \(H^+\) ions changed to \(1\times 10^{-4} M\)
\([H^+]'=1\times 10^{-4} M\)
The pH of the solution will changes to:
\(pH=-\log[H^+]'=-\log[1\times 10^{-4}]=4\)
pH+pOH = 14
pOH = 14 - 4 = 10
And we know that :
\(pOH=-\log[OH^-]\)
\(10=-\log[OH^-]\)
\([OH^-]=1\times 10^{-10} M\)
The new concentration of hydroxide ions is \(1\times 10^{-10} M\).
