25 mL of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a titre value of 35 mL. The molarity of barium hydroxide solution was

Question

25 mL of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a titre value of 35 mL. The molarity of barium hydroxide solution was (A) 0.07 (B) 0.14 (C) 0.28 (D) 0.35

Tardigrade Admin · Accepted Answer

Ba ( OH )2+2 HCl longrightarrow BaCl 2+2 H 2 O (M1 V1/n1)=(M2 V2/r2) [ Ba ( OH )2][H C l] (M1 × 25/1)=(0.1 × 35/2) M1=(0.1 × 35/2 × 25)=0.07 M

Q. 25 mL of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a titre value of 35 mL. The molarity of barium hydroxide solution was

0.07

0.14

0.28

0.35

$B a (O H)_{2} + 2 H Cl ⟶ B a C l_{2} + 2 H_{2} O$
$\frac{M _{1} V _{1}}{n _{1}} = \frac{M _{2} V _{2}}{r _{2}}$
$[B a (O H)_{2}] [H Cl]$
$\frac{M _{1} \times 25}{1} = \frac{0.1 \times 35}{2}$
$M_{1} = \frac{0.1 \times 35}{2 \times 25} = 0.07 M$

Q. 25 mL of a solution of barium hydroxide on titration with a 0.1 molar solution of hydrochloric acid gave a titre value of 35 mL. The molarity of barium hydroxide solution was

0.07

0.14

0.28

0.35

Ba(OH)2​+2HCl⟶BaCl2​+2H2​O n1​M1​V1​​=r2​M2​V2​​ [Ba(OH)2​][HCl] 1M1​×25​=20.1×35​ M1​=2×250.1×35​=0.07M

$B a (O H)_{2} + 2 H Cl ⟶ B a C l_{2} + 2 H_{2} O$
$\frac{M _{1} V _{1}}{n _{1}} = \frac{M _{2} V _{2}}{r _{2}}$
$[B a (O H)_{2}] [H Cl]$
$\frac{M _{1} \times 25}{1} = \frac{0.1 \times 35}{2}$
$M_{1} = \frac{0.1 \times 35}{2 \times 25} = 0.07 M$