2.5 ml of (2/5) M weak monoacidic base (Kp=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 (Kw =1 × 10-14. at .25° C)

Question

2.5 ml of (2/5) M weak monoacidic base (Kp=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 (Kw =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

m mol of base =2.5 × (2/5)=1 m mol of acid required to reach the end point =1 Volume of acid required to reach the end point =(15/2) Total volume at the end point =(15/2)+2.5=10 ml. Molarity of salt at the end point =(1/10)= 0.10 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/a1f767c956bafe8e2286a7ef0eca54c4-.png" /> Kh=(Kw/Kb)=10-2 Kh = 10-2 = (Cα2/- -α) = (0.1 α2/1 - α) ⇒ 10α2+α-1=0 ⇒ α=(-1+√1+40/20)=0.27 ⇒[H+]=C α=0.1 × 0.27=0.027 M

Q. $2.5 m l$ of $\frac{2}{5} M$ weak monoacidic base $(K_{p} = 1 \times 1 0^{- 12}$ 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 (Kp​=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

Q. $2.5 m l$ of $\frac{2}{5} M$ weak monoacidic base $(K_{p} = 1 \times 1 0^{- 12}$ 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 $2 5^{\circ} C)$

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

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

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

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