50 mL water is mixed with 50 mL 0.01 M Ba( | vedclass

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

Question

(A) Step $1$: Calculate the total volume of the solution. $V_{total} = 50 \ mL + 50 \ mL = 100 \ mL = 0.1 \ L$. Step $2$: Calculate the moles of $Ba(O

Vedclass · Accepted Answer

$12.0$

Vedclass · Answer

(A) Step $1$: Calculate the total volume of the solution. $V_{total} = 50 \ mL + 50 \ mL = 100 \ mL = 0.1 \ L$. Step $2$: Calculate the moles of $Ba(OH)_2$. $n = M 	imes V = 0.01 \ mol/L 	imes 0.05 \ L = 5 	imes 10^{-4} \ mol$. Step $3$: Calculate the concentration of $OH^-$ ions. Since $Ba(OH)_2 ightarrow Ba^{2+} + 2OH^-$,the moles of $OH^-$ are $2 	imes (5 	imes 10^{-4}) = 10^{-3} \ mol$. $[OH^-] = \frac{10^{-3} \ mol}{0.1 \ L} = 0.01 \ M = 10^{-2} \ M$. Step $4$: Calculate $pOH$. $pOH = -\log[OH^-] = -\log(10^{-2}) = 2$. Step $5$: Calculate $pH$. $pH = 14 - pOH = 14 - 2 = 12.0$.

$50 \ mL$ water is mixed with $50 \ mL$ $0.01 \ M$ $Ba(OH)_2$. Calculate its $pH$.

Similar Questions

The $pH$ value of which one of the following is not equal to $1$?

An analyst wants to convert $1 \ L$ of $HCl$ solution with $pH = 1$ to a solution of $HCl$ with $pH = 2$. The volume of water needed to do this dilution is $...... \ mL$. (Nearest Integer)

If $0.4 \ g$ of $NaOH$ is present in $1 \ L$ of solution,then the $pH$ of the solution is .....

$0.023 \ g$ of sodium metal is reacted with $100 \ cm^{3}$ of water. The $pH$ of the resulting solution is

$40 \ mg$ of pure sodium hydroxide is dissolved in $10 \ L$ of distilled water. The $p^H$ of the solution is ..........

Vedclass Test Series

Exam Paper Generator

Online Exam Module