用0.1mol/L氢氧化钠溶液滴定100ml0.1mol/L盐酸时,如果滴定误差在正负0.1% 用0.1mol/L的NaOH溶液滴定100ml/L盐酸时，如...
\u75280.10mol.L-1NaOH \u6eb6\u6db2\u6ef4\u5b9a100mL0.10mol.L-1\u76d0\u9178\u65f6\uff0c\u5982\u679c\u6ef4\u5b9a\u8bef\u5dee\u5728\u6b63\u8d1f0.1%\u4ee5\u5185\uff0c\u53cd\u5e94\u5b8c\u6bd5\u540e\u6eb6\u6db2\u7684PH\u56e0\u4e3a\u7406\u8bba\u4e0a\u9700\u8981100mL 0.10mol/L 1NaOH \u6eb6\u6db2,\u8bef\u5dee\u8303\u56f4\u57280.1%\u5185\uff0c\u5373\u00b10.1mL
1)\u5c11\u52a00.1mLNaOH\u65f6\uff0c\u76d0\u9178\u8fc7\u91cf0.1mL
c(H+)=(0.10mol/L*0.1mL )/(100+99.9)mL=5*10^-5mol/L
pH=-lg5*10^-5=4.3
2)\u591a\u52a00.1mLNaOH\u65f6\uff0c\u6c22\u6c27\u5316\u94a0\u8fc7\u91cf0.1mL
c(OH-)=(0.10mol/L*0.1mL )/(100.1+100)mL=5*10^-5mol/L
c(H+)=Kw /c(OH-)=1.0*10^-14 /5*10^-5=2*10^-10
pH=-lg2*10^-10=9.7
3)\u82e5\u4e24\u4e2a\u8bd5\u5242\u4e2d\u5176\u4e2d\u4e00\u4e2a\u4e3a\u5f31\u9178\u6216\u5f31\u78b1\uff0c
\u5047\u8bbe:\u918b\u9178\u4e0e\u6c22\u6c27\u5316\u94a0
1)\u5c11\u52a00.1mLNaOH\u65f6\uff0c\u918b\u9178\u8fc7\u91cf0.1mL,
\u6b64\u65f6\u5f97\u5230\u7f13\u51b2\u6eb6\u6db2
pH=pKa + lgc(Ac-)/c(HAc)=4.76+lg100/0.1=7.76
\u6216\u901a\u8fc7:
Ka(HAc)=[H+]*[Ac-]/[HAc]=\u7ea6=[H+]*n(Ac-)/n(HAc)=[H+]*100mL/0.1mL
c(H+)=1.75*10^-5/10^3=1.75*10^-8
pH=-lg1.75*10^-8=7.76
2)\u591a\u52a00.1mLNaOH\u65f6\uff0c\u6c22\u6c27\u5316\u94a0\u8fc7\u91cf0.1mL
\u6b64\u65f6\u6eb6\u6db2\u4e3a\u6c22\u6c27\u5316\u94a0\u548c\u918b\u9178\u94a0\u6eb6\u6db2,\u4e00\u822c\u8ba4\u4e3a
\u918b\u9178\u94a0\u6eb6\u6db2\u96be\u4ee5\u6c34\u89e3
c(OH-)=(0.10mol/L*0.1mL )/(100.1+100)mL=5*10^-5mol/L
c(H+)=Kw /c(OH-)=1.0*10^-14 /5*10^-5=2*10^-10
pH=-lg2*10^-10=9.7
\u6240\u5f97\u6eb6\u6db2PH\u7684\u8303\u56f4\uff1a7.7~9.7
\u663e\u7136\u8303\u56f4\u7f29\u5c0f
C \u6b630.1%\uff1a\u6eb6\u6db2\u4e2dOH-\u7684\u6d53\u5ea6C=0.1mol/L*(0.1%*100mL)/(100mL+100mL)=5*10^(-5)mol/L
pOH=-lg5*10^(-5)=5-lg5=4.3
pH=14-4.3=9.7
\u8d1f0.1%\uff1a\u6c22\u6c27\u5316\u94a0\u91cf\u4e0d\u8db3\uff0c\u76d0\u9178\u6709\u5269\u4f59\uff0c\u5269\u4f59\u91cfn=0.1mol/L*0.1L-0.1mol/L*[0.1L*(1-0.1%)=0.00001mol
C(H+)=N/(0.1L+0.1L)=5*10^(-5)mol/L
\u6240\u4ee5pH=-lgC(H+)=4.3
pOH=-lg5*10^(-5)=5-lg5=4.3
pH=14-4.3=9.7
负0.1%：氢氧化钠量不足，盐酸有剩余，剩余量n=0.1mol/L*0.1L-0.1mol/L*[0.1L*(1-0.1%)=0.00001mol
C(H+)=N/(0.1L+0.1L)=5*10^(-5)mol/L
所以pH=-lgC(H+)=4.3
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