Millimoles of calcium hydroxide required to p | vedclass

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

Question

(A) Given $pH = 12$,we know that $pOH = 14 - pH = 14 - 12 = 2$. $[OH^-] = 10^{-pOH} = 10^{-2} \ M$. Since $Ca(OH)_2$ dissociates completely as $Ca(OH)

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) Given $pH = 12$,we know that $pOH = 14 - pH = 14 - 12 = 2$. $[OH^-] = 10^{-pOH} = 10^{-2} \ M$. Since $Ca(OH)_2$ dissociates completely as $Ca(OH)_2 ightarrow Ca^{2+} + 2OH^-$,the concentration of $Ca(OH)_2$ is $[Ca(OH)_2] = \frac{[OH^-]}{2} = \frac{10^{-2}}{2} = 0.5 	imes 10^{-2} = 5 	imes 10^{-3} \ M$. Number of millimoles $= 	ext{Molarity} 	imes 	ext{Volume in } mL = (5 	imes 10^{-3} \ M) 	imes (100 \ mL) = 0.5 \ 	ext{millimoles}$. This can be written as $5 	imes 10^{-1} \ 	ext{millimoles}$. Comparing with $x 	imes 10^{-1}$,we get $x = 5$.

Millimoles of calcium hydroxide required to produce $100 \ mL$ of the aqueous solution of $pH$ $12$ is $x \times 10^{-1}$. The value of $x$ is $..............$. (Nearest integer). Assume complete dissociation.

Similar Questions

Which of the following solutions will have $pH$ close to $1.0$?

Given that the dissociation constant for $H_2O$ is $K_w = 1 \times 10^{-14} \ mol^2/L^2$. What is the $pH$ of a $0.001 \ M$ $KOH$ solution?

In a solution of $pH = 5$,more acid is added in order to reduce the $pH$ to $2$. The increase in hydrogen ion concentration is $......$ times.

How many moles of $HCl$ must be removed from $1 \ L$ of an aqueous $HCl$ solution to change its $pH$ from $2$ to $3$?

The $pH$ of $10^{-8} \ M$ aqueous solution of $HCl$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module