The concentration of [ H +]and concentration of [ OH -]of a 0.1 aqueous solution of 2 % ionised weak acid is: [lonic product of water =1 × 10∼ 14 ]

Question

The concentration of [ H +]and concentration of [ OH -]of a 0.1 aqueous solution of 2 % ionised weak acid is: [lonic product of water =1 × 10∼ 14 ] (A) 2 × 10-3 M and 5 × 10-12 M (B) 1 × 103 M and 3 × 10-11 M (C) 0.02 × 10-3 M and 5 × 10-11 M (D) 3 × 10-2 M and 4 × 10-13 M

Tardigrade Admin · Accepted Answer

Use the following formula to find concentration of

O H^{-}

ions.

[H^{+}] = C α

[O H^{-}] = \frac{ionic product of water}{[ H ^{+} ]}

Given,

C = 0.1

α = 2

[H^{+}] = C α

= 0.1 \times 0.01

= 2 \times 1 0^{- 3}

[O H^{-}] = \frac{1 \times 1 0 ^{- 14}}{2 \times 1 0 ^{- 3}}

= 5 \times 1 0^{- 12}

Q. The concentration of $[H^{+}]$ and concentration of $[O H^{-}]$ of a $0.1$ aqueous solution of $2%$ ionised weak acid is:
[lonic product of water $= 1 \times 1 0^{\sim} 14$ ]

$2 \times 1 0^{- 3} M$ and $5 \times 1 0^{- 12} M$

$1 \times 1 0^{3} M$ and $3 \times 1 0^{- 11} M$

$0.02 \times 1 0^{- 3} M$ and $5 \times 1 0^{- 11} M$

$3 \times 1 0^{- 2} M$ and $4 \times 1 0^{- 13} M$

Q. The concentration of [H+]and concentration of [OH−]of a 0.1 aqueous solution of 2% ionised weak acid is: [lonic product of water =1×10∼14 ]

2×10−3M and 5×10−12M

1×103M and 3×10−11M

0.02×10−3M and 5×10−11M

3×10−2M and 4×10−13M

Use the following formula to find concentration of OH− ions. [H+]=Cα [OH−]=[H+] ionic product of water ​ Given, C=0.1 α=2 [H+]=Cα =0.1×0.01 =2×10−3 [OH−]=2×10−31×10−14​ =5×10−12

Q. The concentration of $[H^{+}]$ and concentration of $[O H^{-}]$ of a $0.1$ aqueous solution of $2%$ ionised weak acid is:
[lonic product of water $= 1 \times 1 0^{\sim} 14$ ]

$2 \times 1 0^{- 3} M$ and $5 \times 1 0^{- 12} M$

$1 \times 1 0^{3} M$ and $3 \times 1 0^{- 11} M$

$0.02 \times 1 0^{- 3} M$ and $5 \times 1 0^{- 11} M$

$3 \times 1 0^{- 2} M$ and $4 \times 1 0^{- 13} M$