What is the pOH of a 1 mM solution of Ca(OH) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The concentration of the solution is $1 \ mM = 10^{-3} \ M$. Since $Ca(OH)_2$ is a strong base,it dissociates completely as: $Ca(OH)_2 \rightarrow

Vedclass · Accepted Answer

$2.7$

Vedclass · Answer

(A) The concentration of the solution is $1 \ mM = 10^{-3} \ M$. Since $Ca(OH)_2$ is a strong base,it dissociates completely as: $Ca(OH)_2 ightarrow Ca^{2+} + 2OH^-$. For every $1 	ext{ mole}$ of $Ca(OH)_2$,$2 	ext{ moles}$ of $OH^-$ ions are produced. Therefore,$[OH^-] = 2 	imes 10^{-3} \ M$. $pOH = -\log_{10} [OH^-] = -\log_{10} (2 	imes 10^{-3})$. $pOH = -(\log_{10} 2 + \log_{10} 10^{-3}) = -(0.301 - 3) = 2.699 \approx 2.7$.

What is the $pOH$ of a $1 \ mM$ solution of $Ca(OH)_2$?

Similar Questions

What will be the $pH$ of a $10^{-8} \ M \ HCl$ solution?

The $pH$ of a $0.005 \ M \ H_2SO_4$ solution will be:

Which of the following substances has the highest $pH$ at $25^{\circ} C$?

Equal volumes of two solutions $A$ and $B$ of a strong acid having $pH=6.0$ and $pH=4.0$ respectively are mixed together to form a new solution. The $pH$ of the new solution will be in the range

The concentration of $NaOH$ solution is $10^{-8} \ M$. Find out the $[OH^{-}]$ concentration.

Vedclass Test Series

Exam Paper Generator

Online Exam Module