A solution of 0.1 M weak base (B) is titrate | vedclass

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

Question

(C) From the titration curve,the equivalence point is reached at $V = 6 \ mL$ of $HA$ added.At the half-equivalence point,$V = 3 \ mL$,the concentrati

Vedclass · Accepted Answer

$2.85$

Vedclass · Answer

(C) From the titration curve,the equivalence point is reached at $V = 6 \ mL$ of $HA$ added.At the half-equivalence point,$V = 3 \ mL$,the concentration of the weak base $[B]$ is equal to the concentration of its conjugate acid $[BH^{+}]$.According to the Henderson-Hasselbalch equation for a weak base:$pOH = pK_{b} + \log\left(\frac{[BH^{+}]}{[B]}\right)$At the half-equivalence point,$[BH^{+}] = [B]$,so $\log(1) = 0$.Therefore,$pOH = pK_{b}$.From the graph,at $V = 3 \ mL$,the $pH$ is approximately $11.15$.Since $pH + pOH = 14$ at $25^{\circ}C$,we have $pOH = 14 - 11.15 = 2.85$.Thus,$pK_{b} = 2.85$.

Similar Questions

$2.5 \ mL$ of $\frac{2}{5} \ M$ weak monoacidic base ($K_{b} = 1 \times 10^{-12}$ at $25^{\circ} C$) is titrated with $\frac{2}{15} \ M \ HCl$ in water at $25^{\circ} C$. The concentration of $H^{+}$ at equivalence point is ($K_W = 1 \times 10^{-14}$ at $25^{\circ} C$).

At $27^{\circ}C$,$100 \ mL$ of $0.4 \ M$ $HCl$ is mixed with $100 \ mL$ of $0.5 \ M$ $NaOH$ solution. To the resultant solution,$800 \ mL$ of distilled water is added. What is the $pH$ of the final solution?

Statement $A$: Addition of $NH_4OH$ in the presence of excess $NH_4Cl$ to an aqueous solution of $BaCl_2$ results in the precipitation of $Ba(OH)_2$.
Reason $B$: $Ba(OH)_2$ is insoluble in water.

When $10 \ mL$ of $0.1 \ M$ acetic acid $(pK_a = 5.0)$ is titrated with $10 \ mL$ of $0.1 \ M$ ammonia $(pK_b = 5.0)$,at what $pH$ will the equivalence point be reached?

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module