Question 11.6: A vinegar solution has a [OH^-] = 5.0 × 10^-12 Mat 25 °C. Wh...
A vinegar solution has a [OH^{-}] = 5.0 \times 10^{-12} M at 25 °C. What is the [H_{3}O^{+}] of the vinegar solution? Is the solution acidic, basic, or neutral?
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STEP 1 State the given and needed quantities.
ANALYZE THE PROBLEM | Given | Need | Connect |
[OH^{-}] = 5.0\times10^{-12} M | [H_{3}O^{+}] | K_{w}= [H_{3}O^{+}] [OH^{-}] |
STEP 2 Write the K_{w} for water and solve for the unknown [H_{3}O^{+}].
K_{w}= \boxed{[H_{3}O^{+}]} [OH^{-}] = 1.0 \times 10^{-14}Solve for [H_{3}O^{+}] by dividing both sides by [OH^{-}].
\frac{K_{w} }{[OH^{-}]} = \frac{\boxed{[H_{3}O^{+}]}[\cancel{OH^{-}}]}{[\cancel{OH^{-}}]}\boxed{[H_{3}O^{+}]} =\frac{1.0 \times 10^{-14}}{[OH^{-}]}
STEP 3 Substitute the known [OH^{-}] into the equation and calculate.
\boxed{[H_{3}O^{+}]} =\frac{1.0 \times 10^{-14}}{[5.0 \times 10^{-12}]} = 2.0 \times 10^{-3} MBecause the [H_{3}O^{+}] of 2.0 \times 10^{-3} M is larger than the [OH^{-}] of 5.0 \times 10^{-12} M, the solution is acidic.
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