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    【题目】在25℃时，某稀溶液中由水电离产生的c(OH－)＝10－10mol/L。下列有关该溶液的叙述正确的是( )

    A．该溶液一定呈酸性

    B．该溶液中的c(H＋)可能等于10－5

    C．该溶液的pH可能为4，可能为10

    D．该溶液有可能呈中性

    【答案】C

    【解析】由水电离出来的c(OH－)＝10－10，故溶液中的c(H＋)水＝c(OH－)水＝1×10－10mol/L，若溶液中的c(OH－)＝10－10，则该溶液中的c(H＋)应等于10－4，则溶液的pH应为4，溶液呈酸性；若溶液中的c(H＋)＝10－10，则溶液的pH应为10，(该溶液中的c(OH－)应等于10－4)，溶液呈碱性；溶液可能是呈酸性也可能是呈碱性，但溶液不可能是pH＝5的溶液，当然也不可能呈中性。

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