Calculate the pH of a 0.37 g Ca(OH)_2 soluti | vedclass

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

Question

(A) $1$. Calculate the molar mass of $Ca(OH)_2$: $40 + 2 	imes (16 + 1) = 40 + 34 = 74 \ g/mol$. $2$. Calculate the number of moles of $Ca(OH)_2$: $n

Vedclass · Accepted Answer

$12.6$

Vedclass · Answer

(A) $1$. Calculate the molar mass of $Ca(OH)_2$: $40 + 2 	imes (16 + 1) = 40 + 34 = 74 \ g/mol$. $2$. Calculate the number of moles of $Ca(OH)_2$: $n = \frac{0.37 \ g}{74 \ g/mol} = 0.005 \ mol$. $3$. Calculate the molarity of $Ca(OH)_2$: $M = \frac{0.005 \ mol}{0.2 \ L} = 0.025 \ M$. $4$. Since $Ca(OH)_2$ is a strong base,it dissociates as $Ca(OH)_2 ightarrow Ca^{2+} + 2OH^-$. $5$. The concentration of $OH^-$ ions is $[OH^-] = 2 	imes 0.025 \ M = 0.05 \ M$. $6$. Calculate $pOH$: $pOH = -\log[OH^-] = -\log(0.05) = -\log(5 	imes 10^{-2}) = 2 - \log(5) \approx 2 - 0.699 = 1.301$. $7$. Calculate $pH$: $pH = 14 - pOH = 14 - 1.301 = 12.699 \approx 12.7$.

Calculate the $pH$ of a $0.37 \ g$ $Ca(OH)_2$ solution in $200 \ mL$ of water. (Atomic masses: $Ca = 40, O = 16, H = 1$)

Similar Questions

The ratio of $pH$ of $0.1 \, M \, H_2SO_4$ and that of $0.1 \, N \, H_2SO_4$ is

The number of moles of $Ca(OH)_2$ required to prepare $250 \ mL$ of solution with $pH$ $14$ (assuming complete ionization) is :-

Which would decrease the $pH$ of $25 \ cm^3$ of a $0.01 \ M$ solution of hydrochloric acid?

Given a $0.1 \ M$ solution of each of the following. Which solution has the lowest $pH$?

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module