The moles of H^{+} from H_2O alone in a 1 L, | vedclass

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

Question

(C) Let the concentration of $HCl$ be $C = \sqrt{5} 	imes 10^{-7} \ M = 2.23 	imes 10^{-7} \ M$. Since $HCl$ is a strong acid,it dissociates complet

Vedclass · Accepted Answer

$3.85 	imes 10^{-8}$

Vedclass · Answer

(C) Let the concentration of $HCl$ be $C = \sqrt{5} 	imes 10^{-7} \ M = 2.23 	imes 10^{-7} \ M$. Since $HCl$ is a strong acid,it dissociates completely: $[H^{+}]_{HCl} = C = 2.23 	imes 10^{-7} \ M$. Let $x$ be the concentration of $H^{+}$ ions produced by the dissociation of $H_2O$. The total concentration of $H^{+}$ ions is $[H^{+}]_{total} = [H^{+}]_{HCl} + [H^{+}]_{H_2O} = C + x$. The concentration of $OH^{-}$ ions is $[OH^{-}] = x$ (since $H_2O$ produces equal moles of $H^{+}$ and $OH^{-}$). At $25 \ ^\circ C$,the ionic product of water is $K_w = [H^{+}][OH^{-}] = 10^{-14}$. Substituting the values: $(C + x)(x) = 10^{-14}$. $x^2 + Cx - 10^{-14} = 0$. Using the quadratic formula $x = \frac{-C + \sqrt{C^2 + 4K_w}}{2}$: $x = \frac{-2.23 	imes 10^{-7} + \sqrt{(2.23 	imes 10^{-7})^2 + 4 	imes 10^{-14}}}{2}$. $x = \frac{-2.23 	imes 10^{-7} + \sqrt{4.97 	imes 10^{-14} + 4 	imes 10^{-14}}}{2} = \frac{-2.23 	imes 10^{-7} + \sqrt{8.97 	imes 10^{-14}}}{2}$. $x = \frac{-2.23 	imes 10^{-7} + 2.995 	imes 10^{-7}}{2} = \frac{0.765 	imes 10^{-7}}{2} = 0.3825 	imes 10^{-7} = 3.825 	imes 10^{-8} \ M$. Since the volume is $1 \ L$,the moles of $H^{+}$ from $H_2O$ is $3.825 	imes 10^{-8} \ mol$,which is approximately $3.85 	imes 10^{-8} \ mol$.

The moles of $H^{+}$ from $H_2O$ alone in a $1 \ L$,$\sqrt{5} \times 10^{-7} \ M$ $HCl$ solution at $25 \ ^\circ C$ is ( $\sqrt{5} = 2.23$ )

Similar Questions

Assign $A$,$B$,$C$,$D$ from the given type of reaction.
$2NaOH + Zn(OH)_2 \rightarrow Na_2ZnO_2 + 2H_2O$

The $pH$ of a solution obtained by mixing $50 \, mL$ of $1 \, M \, HCl$ and $30 \, mL$ of $1 \, M \, NaOH$ is $x \times 10^{-4}$. The value of $x$ is ...... . (Nearest integer) $[\log 2.5 = 0.3979]$

Given,$pH$ of a solution $A$ is $3$ and it is mixed with another solution $B$ having $pH$ $2$. If both are mixed in equal volume,then the resultant $pH$ of the solution will be

The $pH$ of the solution containing $10 \ mL$ of $0.1 \ N$ $NaOH$ and $10 \ mL$ of $0.05 \ N$ $H_2SO_4$ would be

The dissociation constant of a certain weak acid is $1.0 \times 10^{-4}$. What is the equilibrium constant for its reaction with a strong base?

Vedclass Test Series

Exam Paper Generator

Online Exam Module