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nevsk [136]
3 years ago
10

Determine whether the following statement is TRUE (T) or FALSE (F).

Chemistry
1 answer:
Alenkasestr [34]3 years ago
4 0

Answer:

I say the statement is true

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What is the pOH of a<br> 2.6 x 10-6 M H+ solution?
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Answer:

Approximately 8.41 (assuming that the solution is at \rm 25^\circ C, under which K_{\rm w} = 10^{-14}.)

Explanation:

Let {\rm [H^{+}]} and {\rm [OH^{-}]} denote the concentration of \rm H^{+} and \rm OH^{-} respectively.

Let K_{\rm w} denote the self-ionization constant of water. The exact value of K_{\rm w}\! depends on the temperature of the solution. K_{\rm w} =10^{-14} at \rm 25^\circ C.

The product of {\rm [H^{+}]} and {\rm [OH^{-}]} in a solution (with \rm M, or moles per liter, as the unit) is supposed to be equal to the K_{\rm w} value of that solution at the corresponding temperature. In other words:

{\rm [H^{+}]} \cdot {\rm [OH^{-}]} = K_{\rm w}.

Rearrange to obtain an expression for {[\rm OH^{-}]}:

\begin{aligned}{\rm [OH^{-}]} &= \frac{K_{\rm w}}{[\rm H^{+}]}\end{aligned}.

Assume that the solution in this question is at \rm 25^\circ C (for which K_{\rm w} =10^{-14}.) For this solution:

\begin{aligned}{\rm [OH^{-}]} &= \frac{K_{\rm w}}{[\rm H^{+}]} \\ &= \frac{10^{-14}}{2.6 \times 10^{-6}}\approx 3.85\times 10^{-9}\; \rm M\end{aligned}.

Hence, the \rm pOH of this solution would be:

\begin{aligned}\rm pOH &= -\log_{10}{\rm [OH^{-}]} \\&\approx -\log_{10} (3.85 \times 10^{-9}) \approx 8.41 \end{aligned}.

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are all physical properties

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