the ph of a solution decreases by 2.0. how does the hydronium ion concentration of the solution change?\no…

the ph of a solution decreases by 2.0. how does the hydronium ion concentration of the solution change?\no increases to 2 times the original concentration\no increases to 100 times the original concentration\no decreases to $\frac{1}{100}$ of the original concentration\no decreases to $\frac{1}{2}$ of the original concentration

the ph of a solution decreases by 2.0. how does the hydronium ion concentration of the solution change?\no increases to 2 times the original concentration\no increases to 100 times the original concentration\no decreases to $\frac{1}{100}$ of the original concentration\no decreases to $\frac{1}{2}$ of the original concentration

Answer

Explanation:

Step1: Recall pH - hydronium ion concentration formula

The formula for pH is $pH = -\log[H_3O^+]$, where $[H_3O^+]$ is the hydronium - ion concentration. Let the initial pH be $pH_1$ and the final pH be $pH_2$. We know that $pH_1−pH_2 = 2.0$.

Step2: Express pH in terms of hydronium ion concentration

$pH_1=-\log[H_3O^+]_1$ and $pH_2 = -\log[H_3O^+]_2$. Then $(-\log[H_3O^+]_1)-(-\log[H_3O^+]_2)=2.0$.

Step3: Simplify the logarithmic equation

Using the property of logarithms $\log a-\log b=\log\frac{a}{b}$, we get $\log\frac{[H_3O^+]_2}{[H_3O^+]_1}=2.0$.

Step4: Solve for the ratio of hydronium ion concentrations

If $\log\frac{[H_3O^+]_2}{[H_3O^+]_1}=2.0$, then by the definition of logarithms (if $\log x = y$, then $x = 10^y$), we have $\frac{[H_3O^+]_2}{[H_3O^+]_1}=10^{2.0}=100$. So $[H_3O^+]_2 = 100\times[H_3O^+]_1$.

Answer:

increases to 100 times the original concentration