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Question-31619




Question Number 31619 by Tinkutara last updated on 11/Mar/18
Answered by Joel578 last updated on 11/Mar/18
Cu(OH)_2  + H_2 SO_4  → CuSO_4  + 2H_2 O  ⇒ weak acid + strong acid → acid salt, pH < 7    CH_3 COOH + NaOH → CH_3 COONa + H_2 O  ⇒ weak acid + strong base → base salt, pH > 7    KOH + HNO_3  → KNO_3  + H_2 O  ⇒ strong base + strong acid → neutral salt, pH = 7
$$\mathrm{Cu}\left(\mathrm{OH}\right)_{\mathrm{2}} \:+\:\mathrm{H}_{\mathrm{2}} \mathrm{SO}_{\mathrm{4}} \:\rightarrow\:\mathrm{CuSO}_{\mathrm{4}} \:+\:\mathrm{2H}_{\mathrm{2}} \mathrm{O} \\ $$$$\Rightarrow\:\mathrm{weak}\:\mathrm{acid}\:+\:\mathrm{strong}\:\mathrm{acid}\:\rightarrow\:\mathrm{acid}\:\mathrm{salt},\:\mathrm{pH}\:<\:\mathrm{7} \\ $$$$ \\ $$$$\mathrm{CH}_{\mathrm{3}} \mathrm{COOH}\:+\:\mathrm{NaOH}\:\rightarrow\:\mathrm{CH}_{\mathrm{3}} \mathrm{COONa}\:+\:\mathrm{H}_{\mathrm{2}} \mathrm{O} \\ $$$$\Rightarrow\:\mathrm{weak}\:\mathrm{acid}\:+\:\mathrm{strong}\:\mathrm{base}\:\rightarrow\:\mathrm{base}\:\mathrm{salt},\:\mathrm{pH}\:>\:\mathrm{7} \\ $$$$ \\ $$$$\mathrm{KOH}\:+\:\mathrm{HNO}_{\mathrm{3}} \:\rightarrow\:\mathrm{KNO}_{\mathrm{3}} \:+\:\mathrm{H}_{\mathrm{2}} \mathrm{O} \\ $$$$\Rightarrow\:\mathrm{strong}\:\mathrm{base}\:+\:\mathrm{strong}\:\mathrm{acid}\:\rightarrow\:\mathrm{neutral}\:\mathrm{salt},\:\mathrm{pH}\:=\:\mathrm{7} \\ $$
Commented by Tinkutara last updated on 11/Mar/18
Is it a typological error in B column option c?
Commented by Joel578 last updated on 11/Mar/18
Yes, I think so
$$\mathrm{Yes},\:\mathrm{I}\:\mathrm{think}\:\mathrm{so} \\ $$
Commented by Tinkutara last updated on 11/Mar/18
Thanks Sir! ����☺��

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