Question #091ee

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Question #091ee

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Answer 1 · 2015-05-27T00:04:28

The pH of your solution will be equal to 13.0.

Sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce sodium cations, $N a^{+}$ $Na^(+)$ . and hydroxide anions, $O H^{-}$ $OH^(-)$ , according to the following chemical equation

$N a O H_{(a q)} \to N a_{(a q)}^{+} + O H_{(a q)}^{-}$ $NaOH_((aq)) -> Na_((aq))^(+) + OH_((aq))^(-)$

![http://wps.prenhall.com/wps/media/objects/476/488316/ch14.html]( useruploads.socratic.org )

Notice that 1 mole of sodium hydroxide produces 1 mole of $N a^{+}$ $Na^(+)$ and 1 mole of $O H^{-}$ $OH^(-)$ ions. This means that the concentration of the hydroxide ions will be equal to

$[O H^{-}] = [N a O H] = 0.1 M$ $[OH^(-)] = [NaOH] = "0.1 M"$

You can use this concentration to determine the pOH of the solution

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

$p O H = - log (0.1) = 1.0$ $pOH = -log(0.1) = 1.0$

Therefore, the pH of the solution will be equal to

$p H_{sol} = 14 - p O H = 14 - 1.0 = 13.0$ $pH_"sol" = 14 - pOH = 14 -1.0 = color(green)(13.0)$

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Question #091ee

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