Question 10.6: Calculating the [H3O^+] of a Solution A vinegar solution has...
Calculating the [ \pmb{H_{3} O^{+}} ] of a Solution
A vinegar solution has a \left[OH^{-}\right] = 5.0 \times 10^{-12} M at 25 °C. What is the \left[H_{3} O^{+}\right] of the vinegar solution? Is the solution acidic, basic, or neutral?
Step \fbox{1} State the given and needed quantities.
| Given | Need | Know | |
| ANALYZE THE PROBLEM |
\left[OH^{-}\right] = 5.0 \times 10^{-12} M | \left[H_{3} O^{+}\right] | K_{w}=\left[H_{3} O^{+}\right]\left[OH^{-}\right] |
| =1.0 \times 10^{-14} |
Step \fbox{2} Write the K_{w} for water and solve for the unknown \left[H_{3} O^{+}\right].
Rearrange the ion product expression by dividing both sides by the \left[OH^{-}\right].
\begin{aligned}K_{w} &=\left[H_{3} O^{+}\right]\left[OH^{-}\right]=1.0 \times 10^{-14} \\\frac{K_{w}}{\left[OH^{-}\right]} &=\frac{\left[H_{3} O^{+}\right]\left[\cancel{OH^{-}}\right]}{[\cancel{OH^{-}}]} \\{\left[H_{3} O^{+}\right] } &=\frac{1.0 \times 10^{-14}}{\left[OH^{-}\right]}\end{aligned}Step \fbox{3} Substitute the known \left[OH^{-}\right] into the equation and calculate.
\left[H_{3} O^{+}\right]=\frac{1.0 \times 10^{-14}}{\left[5.0 \times 10^{-12}\right]}=2.0 \times 10^{-3} MBecause the \left[H_{3} O^{+}\right] of 2.0 \times 10^{-3} M is larger than the \left[OH^{-}\right] of 5.0 \times 10^{-12} M, the solution is acidic.
Guide to Calculating [ \pmb{H_{3} O^{+}} ] and [ \pmb{OH^{-}} ] in Aqueous Solutions
Step \fbox{1}
State the given and needed quantities.
Step \fbox{2}
Write the K_{w} for water and solve for the unknown \left[H_{3} O^{+}\right] or \left[OH^{-}\right].
Step \fbox{3}
Substitute the known \left[H_{3} O^{+}\right] or \left[OH^{-}\right] into the equation and calculate.