2.5 ml 0.4(M) weak monoacidic base (kb=1 × 10-12.. at 25° C ) is titrated with (2/15)(M) H C l in water at 25° C. The concentration of H+at equivalence point is (kw=1 × 10-14., at .25° C )

Question

2.5 ml 0.4(M) weak monoacidic base (kb=1 × 10-12.. at 25° C ) is titrated with (2/15)(M) H C l 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

N1 V1=N2 V2( At equivalence point ) 0.4 × 2.5=(2/15) × V2 B O H(a q)+H C l(a q) arrow B C l-(a q)+H2 O(I) ∴ V2=(0.4 × 2.5/(2)15) =(0.4 × 2.5/2) × 15=7.5 ml ∴ Concentration of salt =(0.4 × 2.5/2.5+7.5) or, Salt of weak base and strong acid =0.1 ∴ [H+]=C h=√(Kw C/Kb) =√(10-14 × 0.1/10-12)=√10-3 =3.2 × 10-2 M as C h2=(Kw/Kb) ∴ h=√(Kw/Kb C)

Q. $2.5 m l 0.4 (M)$ 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. 2.5ml0.4(M) weak monoacidic base (kb​=1×10−12. at 25∘C ) is titrated with 152​(M)HCl in water at 25∘C. The concentration of H+at equivalence point is (kw​=1×10−14, at 25∘C)

3.7×10−13(M)

3.2×10−7(M)

3.2×10−2(M)

2.7×10−2(M)

Q. $2.5 m l 0.4 (M)$ 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 $2 5^{\circ} C)$

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

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

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

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