Question

How many times concentrated will be the aqueous solution having pH 11.9 as compared to aqueous solution having pH 8? Solve the given problem.

Answer 1

Final Answer :$1.25 * 10^{-4} M (approx)$

Symbols have desired meaning as per question.
log has base 10 here.
Take $\log (5) = 0.7 , \log(2)= 0.3$

Steps:
1) We have,
$pH_1 = 11.9 \\ pH_2 = 8$

So,
We know pH is simply negative logarithm base 10 of H+ concentration.

$pH_1- pH_2 = 3.9=4-0.1 \\ => -\log({H_1^+ })-(-\log({H_2^+}) = \log(10^4)-[\log(5) -2\log(2) ]\\ => \log(\frac{H_2^+}{H_1^+})= \log(10^4)- \log(\frac{5}{4}) \\ => \log(\frac{H_1^+}{H_2^+})= -\log(10^4)+\log(\frac{5}{4}) \\ => \frac{H_1^+}{H_2^+} = \frac{5}{4} *10^{-4} =1.25 *10^{-4}$

Hence, Concentration(in terms of H+ concentration) of solution having pH = 11.9 is 1.25 *10^(-4) times solution having pH= 8 .

Answer 2

Answer:

hey mate, here is your answer..

Explanation:

hope it helps u frnd !!