Question 8.SP.4: Calculating the [H3O^+] of a Solution A vinegar solution has......
Calculating the [H_{3} O^{+}] of a Solution
A vinegar solution has a [OH^{-}]=5.0 ×10^{-12} M at 25 °C What is the [H_{3} O^{+}]of the vinegar solution? Is the solution acidic, basic, or neutral?
Step-by-Step
Step 1 Write the K_{w} for water.
K_{w}=[H_{3} O^{+}][OH^{-}]= 1.0 ×10^{-14}Step 2 Solve the K_{w} for the unknown [H_{3} O^{+}]. Rearrange the ion product expression by dividing through by the [OH^{-}]
\frac{K_{w}}{[OH^{-}]} =\frac{[H_{3} O^{+}] \cancel{[OH^{-}]}}{\cancel{[OH^{-}]}} .
[H_{3} O^{+}]=\frac{1.0 ×10^{-14}}{[OH^{-}]}
Step 3 Substitute in the known [OH^{-}] and calculate.
[H_{3} O^{+}]=\frac{1.0 ×10^{-14}}{[5.0 ×10^{-12}]}=2.0 ×10^{-3} MBecause the [H_{3} O^{+}] of 2.0 ×10^{-3} M is larger than the [OH^{-}] of 5.0 ×10^{-12} M, the solution is acidic.
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