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Question Number 31619 by Tinkutara last updated on 11/Mar/18

Answered by Joel578 last updated on 11/Mar/18

$$\mathrm{Cu}{\left(\mathrm{OH}\right)}_2+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{CuSO}}_4+{2\mathrm{H}}_2\mathrm{O}$$$$\Rightarrow \mathrm{weak}\mathrm{acid}+\mathrm{strong}\mathrm{acid}\to \mathrm{acid}\mathrm{salt},\mathrm{pH}<7$$$$$$$${\mathrm{CH}}_3\mathrm{COOH}+\mathrm{NaOH}\to {\mathrm{CH}}_3\mathrm{COONa}+{\mathrm{H}}_2\mathrm{O}$$$$\Rightarrow \mathrm{weak}\mathrm{acid}+\mathrm{strong}\mathrm{base}\to \mathrm{base}\mathrm{salt},\mathrm{pH}>7$$$$$$$$\mathrm{KOH}+{\mathrm{HNO}}_3\to {\mathrm{KNO}}_3+{\mathrm{H}}_2\mathrm{O}$$$$\Rightarrow \mathrm{strong}\mathrm{base}+\mathrm{strong}\mathrm{acid}\to \mathrm{neutral}\mathrm{salt},\mathrm{pH}=7$$

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

$$\mathrm{Yes},\mathrm{I}\mathrm{think}\mathrm{so}$$

Commented by Tinkutara last updated on 11/Mar/18

Thanks Sir! ����☺��

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