  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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