Chemical And Ionic Equilibrium Result Question 34
34. At $90^{\circ} \mathrm{C}$, pure water has $\left[\mathrm{H}_3 \mathrm{O}^{+}\right]$ as $10^{-6} \mathrm{molL}^{-1}$. What is the value of $K_w$ at $90^{\circ} \mathrm{C}$ ?
(1981, 1M)
(a) $10^{-6}$
(b) $10^{-12}$
(c) $10^{-14}$
(d) $10^{-8}$
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Answer:
Correct Answer: 34. ( b )
Solution:
- $K_w=\left[\mathrm{H}_3 \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=10^{-6} \times 10^{-6}=10^{-12}$
BETA
