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Question

What is the $$pH$$ of a $$1.4\cdot 10^{−2}M\ NaOH$$ solution?


Solution

$$pH=12.15$$ 
Explanation:
Even before doing any calculations, you can say that since you're dealing with a strong base, the $$pH$$ of the solution must be higher than 7.
The higher the concentration of the base, the higher the pH will be.
In your case, you're dealing with a solution of sodium hydroxide, $$NaOH$$, a strong base that dissociates completely in aqueous solution to form sodium cations, $$Na^+$$, and hydroxide anions, $$OH^-$$
$$NaOH_{(aq]}\to Na^+_{(aq]}+OH^−_{(aq]}$$
Notice that the salt dissociates in a 1:1 mole ratio with the hydroxide anions, you can say that
$$[OH^-]=[NaOH]=1.4\cdot 10^{−2}M$$
Now, the $$pH$$ of the solution is determined by the concentration of hydronium ions, $$H_3O^+$$. For aqueous solutions, the concentration of hydronium ions is related to the concentration of hydroxide ions by the ion product constant of water, $$K_W$$
$$K_W=[OH^−]\cdot [H_3O^+]$$ 
At room temperature, you have
$$K_W=10^{−14}$$
This means that the concentration of hydronium ions can be determined by using
$$[_H3O^+]=\dfrac{K_W}{[OH^−]}$$ 
Plug in your values to get
$$[H_3O^+]$$=$$\dfrac{10^{−14}}{1.4 \times 10^{--2}}$$=$$7.14\cdot 10^{−13}M$$
The $$pH$$ of the solution is equal to
$$pH=−\log([H_3O^+])$$ 
In your case,
$$pH=−\log(7.14\cdot 10^{−13})=12.15$$
As predicted, the pH is not only higher than 7, but it is significantly higher than 7.
Alternatively, you can use the $$pOH$$ of the solution to find its pH. As you know,
$$pOH=−\log([OH^-])$$
In your case,
$$pOH=−\log(1.4⋅10^{-2})=1.85$$
You know that
$$pH+pOH=14$$
and so, once again, you have
$$pH=14-1.85=12.15$$

General Knowledge

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