0.025mol/lBa(oh)2溶液20ml与0.15mol/lNaoH溶液20ml混合求40ml溶液的PH
问题描述:
0.025mol/lBa(oh)2溶液20ml与0.15mol/lNaoH溶液20ml混合求40ml溶液的PH
答
氢氧根离子:[OH-] = (0.025 * 0.02 *2 + 0.15 * 0.02) ÷ 0.04
= 0.004 ÷ 0.04 = 0.1 (mol/L)
则:[H+] = Kw ÷ [OH-] = 10^-14 ÷ 0.1 = 10^-13 (mol/L)
所以:PH = 13