常温下将0.010molCH3COONa和0.004molHCl溶于水,配制成0.5L混合溶液,判断:(1)溶液中有两种微粒的物质的量之和一定等于0.010mol,它们是__________和__________;(2)溶液中n(CH3COO-) + n(OH-) - n(H+) = _
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![常温下将0.010molCH3COONa和0.004molHCl溶于水,配制成0.5L混合溶液,判断:(1)溶液中有两种微粒的物质的量之和一定等于0.010mol,它们是__________和__________;(2)溶液中n(CH3COO-) + n(OH-) - n(H+) = _](/uploads/image/z/12308722-34-2.jpg?t=%E5%B8%B8%E6%B8%A9%E4%B8%8B%E5%B0%860%EF%BC%8E010molCH3COONa%E5%92%8C0%EF%BC%8E004molHCl%E6%BA%B6%E4%BA%8E%E6%B0%B4%2C%E9%85%8D%E5%88%B6%E6%88%900%EF%BC%8E5L%E6%B7%B7%E5%90%88%E6%BA%B6%E6%B6%B2%2C%E5%88%A4%E6%96%AD%EF%BC%9A%EF%BC%881%EF%BC%89%E6%BA%B6%E6%B6%B2%E4%B8%AD%E6%9C%89%E4%B8%A4%E7%A7%8D%E5%BE%AE%E7%B2%92%E7%9A%84%E7%89%A9%E8%B4%A8%E7%9A%84%E9%87%8F%E4%B9%8B%E5%92%8C%E4%B8%80%E5%AE%9A%E7%AD%89%E4%BA%8E0%EF%BC%8E010mol%2C%E5%AE%83%E4%BB%AC%E6%98%AF__________%E5%92%8C__________%EF%BC%9B%EF%BC%882%EF%BC%89%E6%BA%B6%E6%B6%B2%E4%B8%ADn%28CH3COO-%29+%2B+n%28OH-%29+-+n%28H%2B%29+%3D+_)
常温下将0.010molCH3COONa和0.004molHCl溶于水,配制成0.5L混合溶液,判断:(1)溶液中有两种微粒的物质的量之和一定等于0.010mol,它们是__________和__________;(2)溶液中n(CH3COO-) + n(OH-) - n(H+) = _
常温下将0.010molCH3COONa和0.004molHCl溶于水,配制成0.5L混合溶液,判断:
(1)溶液中有两种微粒的物质的量之和一定等于0.010mol,它们是__________和__________;
(2)溶液中n(CH3COO-) + n(OH-) - n(H+) = _____________mol.
常温下将0.010molCH3COONa和0.004molHCl溶于水,配制成0.5L混合溶液,判断:(1)溶液中有两种微粒的物质的量之和一定等于0.010mol,它们是__________和__________;(2)溶液中n(CH3COO-) + n(OH-) - n(H+) = _
(1)溶液中有两种微粒的物质的量之和一定等于0.010mol,它们是__醋酸根离子________和_______醋酸___;
(2)溶液中n(CH3COO-) + n(OH-) - n(H+) = ____0.006_____mol.
1.0.01mol的醋酸钠中有钠离子0.01mol,醋酸根0.01mol,溶于水后,醋酸根会结合少量氢离子生成醋酸,醋酸根+醋酸=没溶于水前醋酸根的量=0.01mol.(物料守恒)
2.根据电荷守恒来做因为溶液是不显电性的,所以溶液中正电荷数=负电荷数,n(H+)+n(Na+)=n(cl-)+n(OH-)+n(CH3COO-),其中钠离子的量是0.01mol,氯离子的量是0.004mol,代入得溶液中n(CH3COO-) + n(OH-) - n(H+) =0.01-0.004=0.006mol