Question

part e: what is the ph of a 7.0 x 10-3 m aqueous solution of ca(oh)2 at 25°c?
round your answer to two places past the decimal.
ph=
type your answer...

Explanation:

Step1: Calculate the concentration of \(OH^-\) ions

Calcium hydroxide \(Ca(OH)_2\) dissociates as \(Ca(OH)_2
ightarrow Ca^{2 +}+2OH^-\). Given \([Ca(OH)_2]=7.0\times10^{-3}M\), then \([OH^-]=2\times7.0\times 10^{-3}M = 1.4\times10^{-2}M\)

Step2: Calculate pOH

The formula for pOH is \(pOH=-\log[OH^-]\). Substituting \([OH^-]=1.4\times10^{-2}M\) into the formula: \(pOH =-\log(1.4\times10^{-2})=-\log(1.4)-\log(10^{-2})\approx - 0.15 + 2=1.85\)

Step3: Calculate pH

We know the relationship \(pH + pOH=14\) at \(25^{\circ}C\). So \(pH=14 - pOH\). Substituting \(pOH = 1.85\) into the formula: \(pH=14 - 1.85 = 12.15\)

Answer:

\(12.15\)