A solution contains 8 , g of NaOH and 4.9 , g | vedclass

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

Question

(B) Step $1$: Calculate the number of moles of each solute.Moles of $NaOH = \frac{8 \, g}{40 \, g/mol} = 0.2 \, mol$.Moles of $H_2SO_4 = \frac{4.9 \,

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(B) Step $1$: Calculate the number of moles of each solute.Moles of $NaOH = \frac{8 \, g}{40 \, g/mol} = 0.2 \, mol$.Moles of $H_2SO_4 = \frac{4.9 \, g}{98 \, g/mol} = 0.05 \, mol$.Step $2$: Calculate the equivalents of $H^+$ and $OH^-$.$OH^-$ from $NaOH = 0.2 \, mol 	imes 1 = 0.2 \, eq$.$H^+$ from $H_2SO_4 = 0.05 \, mol 	imes 2 = 0.1 \, eq$.Step $3$: Determine the excess concentration.Excess $OH^- = 0.2 - 0.1 = 0.1 \, mol/L$.Step $4$: Calculate $pOH$ and $pH$.$pOH = -\log[OH^-] = -\log(0.1) = 1$.$pH = 14 - pOH = 14 - 1 = 13$.

$A$ solution contains $8 \, g$ of $NaOH$ and $4.9 \, g$ of $H_2SO_4$ in $1 \, L$ of solution. What is its $pH$?

Similar Questions

One mole of ammonia was completely absorbed in one litre solution each of $(a)$ $1 \ M \ HCl$,$(b)$ $1 \ M \ CH_{3}COOH$,and $(c)$ $1 \ M \ H_{2}SO_{4}$ at $298 \ K$. The decreasing order for the $pH$ of the resulting solutions is (Given $pK_{b}(NH_{3}) = 4.74$)

What will be the change in acidity if
$(i)$ $CuSO_4$ is added in saturated $(NH_4)_2SO_4$ solution
$(ii)$ $SbF_5$ is added in anhydrous $HF$?

$0.1 \, M$ acetic acid is $1\%$ ionized. If its ionization increases to $10\%$,what will be its new concentration (in $, M$)?

When $100 \, mL$ of $M/10 \, NaOH$ solution and $50 \, mL$ of $M/5 \, HCl$ solution are mixed,the $pH$ of the resulting solution would be:

What is the $pH$ of a solution containing $10 \, mL$ of $0.1 \, M \, NaOH$ and $10 \, mL$ of $0.05 \, M \, H_2SO_4$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module