已知NaHSO4在水中的电离方程式为NaHSO4====Na++H++。某温度下,向pH=6的蒸

已知NaHSO4在水中的电离方程式为NaHSO4====Na++H++。某温度下,向pH=6的蒸馏水中加入NaHSO4晶体,保持温度不变,测得溶液pH为2。对于该溶液下列叙述不正确的是(    )

    A.该温度高于25 ℃

    B.水电离出来的c(H+)=1×10-10 mol·L-1

    C.c(H+)=c(OH-)+c()

    D.该温度下加入等体积pH为12的NaOH溶液可使反应后的溶液恰好呈中性

   

答案

解析:因蒸馏水的pH=6,所以水的离子积常数KW=10-12,则即温度高于25 ℃;水电离出来的c(H+)=c(OH-)=mol·L-1=1×10-10 mol·L-1;由电荷守恒:c(H+)+c(Na+)=c(OH-)+2c(),因电离关系:c(Na+)=c(),故c(H+)=c(OH-)+c();D项中pH为12的NaOH溶液c(OH-)=1 mol·L-1与等体积的pH=2的NaHSO4溶液反应后显碱性,故选D。

    答案:D

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