Question

Calculate the H30*] of an aqueous solution that is 4.76 x 107 M in OH". Enter your answer in scientific notation. Be sure to answer all parts. x 10-7 м

Answer 1

Answer:-

Given:-

molar concentration of OH^- ion i.e [OH^-] = 4.76 $\times$ 10^-7 M

molar concentration of H₃O⁺ ion i.e [H₃O^+-] = ?

As we know that ionic product of water (H₂O) at 25 ⁰C is as follows:-

H₂O $\rightarrow$ H⁺ + OH^-

So

equilibrium constant (K) = [H^+-][OH^-] /  [H₂O]

equilibrium constant (K)$\times$[H₂O] = [H^+-][OH^-]

equilibrium constant (K)$\times$[H₂O] = K_w = [H^+-][OH^-] (since [H₂O] = 1) ---------------------------------------(1)

where

K_w = ionic product of water

The ionic product of water at 25 ⁰C is as given below:-

K_w = 1.0 $\times$ 10^-14 -----------------------------------------------------(2)

from equation no. 1 and 2 we get

[H^+-][OH^-] = 1.0 $\times$ 10^-14 -----------------------------------------------------(3)

So equation no. 3 can be written as follows:-

[H₃O^+-][OH^-] = 1.0 $\times$ 10^-14

therefore

molar concentration of H₃O⁺ ion i.e [H₃O^+-] =  1.0 $\times$ 10^-14 / molar concentration of OH^- ion i.e [OH^-]

molar concentration of H₃O⁺ ion i.e [H₃O^+-] =  1.0 $\times$ 10^-14 / 4.76 $\times$ 10^-7

molar concentration of H₃O⁺ ion i.e [H₃O^+-] = 0.21 $\times$ 10^-7 M (i.e the answer)