Question

ph of resultant solution formed by mixing of equal volumes of two solutions having ph 5.8 and ph 12.8

Answer 1

first of all, we have to find out concentration of hydrogen ions in both solutions.

we know from Arrhenius theory,

$pH=-log[H^+]$

5.8 = -log[H^+]

log[H^+] = -5.8

[H^+] = antilog( -5.8) = 1.58489 × 10^-6

similarly, pH = 12.8

-log[H^+] = 12.8

[H^+] = antilog(-12.8) = 1.58489 × 10^-13

now, use formula,
$M_1V_1+M_2V_2=M(V_1+V_2)$

here, V1 and V2 are volumes of each solutions
a/c to question, V1 = V2 = V (let)

then, 1.58489 × 10^-6 × V + 1.58489 × 10^-13 × V = M × 2V

(1.58489 × 10^-6 + 1.58489 × 10^-13)/2 = M

M = 7.92445 × 10^-7

now, pH = -log(7.92445 × 10^-7)

= 6.1

Answer 2

pH of resultant solution formed by mixing of equal volumes of two solutions having pH 5.8 and pH 12.8 is 6.12

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

1. pH = 5.8

$pH=-\log [H^+]$

Putting in the values:

$5.8=-\log[H^+]$

$[H^+]=10^{-5.8}=1.5\times 10^{-6}M$

2. pH = 12.8

$pH=-\log [H^+]$

Putting in the values:

$12.8=-\log[H^+]$

$[H^+]=10^{-12.8}=1.5\times 10^{-13}M$

Thus total $[H^+]=\frac{1.5\times 10^{-6}+1.5\times 10^{-13}}{2}=7.5\times 10^{-7}M$

pH of resulting solution , $pH=-\log[H^+]=-\log[7.5\times 10^{-7}]=6.12$

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