Question

Calculate either [H3O+] or [OH−] for each of the solutions at 25 °C. Solution A: [OH−]=1.55×10−7 M Solution A: [H3O+]= M Solution B: [H3O+]=9.43×10−9 M Solution B: [OH−]= M Solution C: [H3O+]=0.000775 M Solution C: [OH−]= M Which of these solutions are basic at 25 °C? Solution C: [H3O+]=0.000775 M Solution A: [OH−]=1.55×10−7 M Solution B: [H3O+]=9.43×10−9 M

Answer 1

Answer: A. $[H_3O^+]=0.64\times 10^{-7}M$

B. $[OH^-]=0.11\times 10^{-5}M$

C. $[OH^-]=\frac{10^{-14}}{0.000775}=1.3\times 10^{-11}M$

Thus solution B is basic in nature.

Explanation:

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

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

pH for acidic solutions is less than 7, for basic solutions it is more than 7 and for neutral solutions it is equal to 7.

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

$pOH=-log[OH^-]$

$pH+pOH=14$

or $[H^+][OH^-]=10^{-14}$

A. $[OH^-]=1.55\times 10^{-7}M$

$[H_3O^+]=\frac{10^{-14}}{1.55\times 10^{-7}}=0.64\times 10^{-7}M$

$pH=-log[H_3O^+]=-log[0.64\times 10^{-7}]=7$

B. $[H_3O^+]=9.43\times 10^{-9}M$

$[OH^-]=\frac{10^{-14}}{9.43\times 10^{-9}}=0.11\times 10^{-5}M$

$pH=-log[H_3O^+]=-log[9.43\times 10^{-9}]=8$

C. $[H_3O^+]=0.000775M$

$[OH^-]=\frac{10^{-14}}{0.000775}=1.3\times 10^{-11}M$

$pH=-log[H_3O^+]=-log[0.000775]=3$

Thus solution B is basic.