Question #d8f04

Question

Question #d8f04

1 Answer

Impact of this question

2150 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-12-05T23:43:56

The answer is $5.5 \cdot 10^{- 14}$ $5.5 * 10^(-14)$ .

$N a O H$ $NaOH$ is a strong base, which means it dissasociates completely into $N a^{+}$ $Na^+$ and $O H^{-}$ $OH^-$ .

$N a O H (a q) + H_{2} O (l) \Leftrightarrow N a^{+} (a q) + O H^{-} (a q)$ $NaOH(aq) + H_2O(l) <=> Na^+ (aq) + OH^(-) (aq)$

Starting from the given mass of $8.6$ $8.6$ grams of $N a O H$ $NaOH$ and knowing that its molar mass is $40 \frac{g}{m o l e}$ $40 g/(m o l e)$ , we get

$n_{N a O H} = \frac{8.6 g}{40 \frac{g}{m o l e}} = 0.215$ $n_(NaOH) = (8.6 g)/(40 g/(m o l e)) = 0.215$ moles.

The molar concentration of $N a O H$ $NaOH$ is

$C = \frac{n}{V} = \frac{0.215 m o l e s}{1.2 L} = 0.18 M$ $C = n/V = (0.215 mo l e s)/(1.2 L) = 0.18 M$

Now, a strong base (and a strong acid for that matter) dissasociates completely, which means that the concentrations of $N a^{+}$ $Na^+$ and $O H^{-}$ $OH^-$ in aqueous solution are equal to the initial concentration of $N a O H$ $NaOH$ , $0.18 M$ $0.18 M$ .

Therefore, we can calculate the $p O H$ $pOH$ using

$p O H = - log ([O H^{-}]) = - log (0.18) = 0.74$ $pOH = -log([OH^-]) = -log(0.18) = 0.74$

We can determine the $p H$ $pH$ by using

pH $= 14 - p O H = 14 - 0.74 = 13.26$ $= 14 - pOH = 14 - 0.74 = 13.26$

This means that the $[H^{+}]$ $[H^+]$ is equal to

$[H^{+}] = 10^{- p H} = 10^{- 13.26} = 5.5 \cdot 10^{- 14}$ $[H^+] = 10^(-pH) = 10^(-13.26) = 5.5 * 10^(-14)$

Science

Math

Humanities

... and beyond

Question #d8f04

1 Answer

Related questions

Impact of this question