Question

Compared to a 1.0-liter aqueous solution with a pH of 7.0, a 1.0-liter aqueous solution with a pH of 5.0 contains10 times more hydronium100 times more hydronium10 times more hydroxide100 times more hydroxide

Answer 1

Answer: The a solution having pH = 7 has 100 times more hydroxide ions than in a solution having pH = 5

Explanation:

pH is defined as the negative logarithm of hydrogen ion concentration present in the solution.

Mathematically,

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

pOH is defined as the negative logarithm of hydroxide ion concentration present in the solution.

Mathematically,

$pOH=-\log[OH^-]$

• When pH = 7

Putting values in above equation, we get:

$7=-\log[H^+]$

$[H^+]=10^{-7}M$

• When pH = 5

Putting values in above equation, we get:

$5=-\log[H^+]$

$[H^+]=10^{-5}M$

Taking the ratio of hydrogen ion for both the pH:

$\frac{[H^+]_{pH=5}}{[H^+]_{pH=7}}=\frac{10^{-5}}{10^{-7}}\\\\\frac{[H^+]_{pH=5}}{[H^+]_{pH=7}}=10^2$

$[H^+]_{pH=5}=100\times [H^+]_{pH=7}$

To calculate the pOH, we use the equation:

$pH+pOH=14$

• When pH = 7

pOH = 14 - 7 = 7

Putting values in above equation, we get:

$7=-\log[OH^-]$

$[OH^-]=10^{-7}M$

• When pH = 5

pOH = 14 - 5 = 9

Putting values in above equation, we get:

$9=-\log[OH^-]$

$[OH^-]=10^{-9}M$

Taking the ratio of hydrogen ion for both the pH:

$\frac{[OH^-]_{pH=7}}{[OH^-]_{pH=5}}=\frac{10^{-7}}{10^{-9}}\\\\\frac{[OH^-]_{pH=7}}{[OH^-]_{pH=5}}=10^2$

$[OH^-]_{pH=7}=100\times [OH^-]_{pH=5}$

Hence, the a solution having pH = 7 has 100 times more hydroxide ions than in a solution having pH = 5