If equal volumes of 0.1 M HBr and 0.1 M KOH are mixed, then which of the following is/are correct about the resulting solution ? (1) [ H 3 O +]=1.0 × 10-7 mol L -1 (2) [ OH -]=1.0 × 10-7 mol L -1 (3) [ K +]=0.05 mol L -1 (4) [ Br -]=0.10 mol L -1

Question

If equal volumes of 0.1 M HBr and 0.1 M KOH are mixed, then which of the following is/are correct about the resulting solution ? (1) [ H 3 O +]=1.0 × 10-7 mol L -1 (2) [ OH -]=1.0 × 10-7 mol L -1 (3) [ K +]=0.05 mol L -1 (4) [ Br -]=0.10 mol L -1 (A) 1, 2 and 3 are correct (B) 1 and 2 are correct (C) 2 and 4 are correct (D) 1 and 3 are correct

Tardigrade Admin · Accepted Answer

HBr + KOH arrow KBr + H 2 O Volume of resulting solution will be doubled and the solution will be neutral. Hence, [K+]=[B r-]=(0.1/2)=0.05 mol L-1 [ H 3 O +]=[ OH -]=10-7 mol L-1

Q. If equal volumes of 0.1MHBr and 0.1MKOH are mixed, then which of the following is/are correct about the resulting solution ? (1) [H3​O+]=1.0×10−7molL−1 (2) [OH−]=1.0×10−7molL−1 (3) [K+]=0.05molL−1 (4) [Br−]=0.10molL−1

1, 2 and 3 are correct

1 and 2 are correct

2 and 4 are correct

1 and 3 are correct

HBr+KOH→KBr+H2​O Volume of resulting solution will be doubled and the solution will be neutral. Hence, [K+]=[Br−]=20.1​=0.05molL−1 [H3​O+]=[OH−]=10−7molL−1

$H B r + K O H \to K B r + H_{2} O$
Volume of resulting solution will be doubled and the solution will be neutral.
Hence, $[K^{+}] = [B r^{-}] = \frac{0.1}{2} = 0.05 m o l L^{- 1}$
$[H_{3} O^{+}] = [O H^{-}] = 1 0^{- 7} m o l L^{- 1}$