2.5 mL of (2/5) M weak monoacidic base ( K b =1 × 10-12. at .25° C ) is titrated with (2/15) M HCl in water at 25° C. The concentration of H +at equivalence point is ( K w =1 × 10-14. at .25° C )

Question

2.5 mL of (2/5) M weak monoacidic base ( K b =1 × 10-12. at .25° C ) is titrated with (2/15) M HCl in water at 25° C. The concentration of H +at equivalence point is ( K w =1 × 10-14. at .25° C ) (A) 3.7 × 10-13 M (B) 3.2 × 10-7 M (C) 3.2 × 10-2 M (D) 2.7 × 10-2 M

Tardigrade Admin · Accepted Answer

BOH + HC1 longrightarrow undersetCBCl + H 2 O undersetC (1- h )B ++ H 2 O leftharpoons undersetChBOH + undersetChH + Volume of HCl used =(2.5 × (2/5)/2 / 15)=7.5 m 1 Concentration of Salt, C =(2.5 × (2/5)/10)=0.1 M ∴ ( Ch 2/1- h )=( K w / K b ) Solving h =0.27 [ H +]= Ch =0.1 × 0.27=2.7 × 10-2 M

Q. $2.5 m L$ of $\frac{2}{5} M$ weak monoacidic base $(K_{b} = 1 \times 1 0^{- 12}$ at $2 5^{\circ} C)$ is titrated with $\frac{2}{15} M H Cl$ in water at $2 5^{\circ} C$ . The concentration of $H^{+}$ at equivalence point is $(K_{w} = 1 \times 1 0^{- 14}$ at $2 5^{\circ} C)$

$3.7 \times 1 0^{- 13} M$

$3.2 \times 1 0^{- 7} M$

$3.2 \times 1 0^{- 2} M$

$2.7 \times 1 0^{- 2} M$

Q. 2.5mL of 52​M weak monoacidic base (Kb​=1×10−12 at 25∘C) is titrated with 152​MHCl in water at 25∘C. The concentration of H+at equivalence point is (Kw​=1×10−14 at 25∘C)

3.7×10−13M

3.2×10−7M

3.2×10−2M

2.7×10−2M

BOH+HC1⟶CBCl​+H2​O C(1−h)B+​+H2​O⇌ChBOH​+ChH+​ Volume of HCl used =2/152.5×52​​=7.5m1 Concentration of Salt, C=102.5×52​​=0.1M ∴1−hCh2​=Kb​Kw​​ Solving h=0.27 [H+]=Ch=0.1×0.27=2.7×10−2M

Q. $2.5 m L$ of $\frac{2}{5} M$ weak monoacidic base $(K_{b} = 1 \times 1 0^{- 12}$ at $2 5^{\circ} C)$ is titrated with $\frac{2}{15} M H Cl$ in water at $2 5^{\circ} C$ . The concentration of $H^{+}$ at equivalence point is $(K_{w} = 1 \times 1 0^{- 14}$ at $2 5^{\circ} C)$

$3.7 \times 1 0^{- 13} M$

$3.2 \times 1 0^{- 7} M$

$3.2 \times 1 0^{- 2} M$

$2.7 \times 1 0^{- 2} M$