Question

質(zhì)量分?jǐn)?shù)是7.3%的稀鹽酸滴入盛有10g 氫氧化鈉溶液（滴有2滴酚酞試液）的燒杯中，恰好完全反應(yīng)時(shí)消耗此稀鹽酸的質(zhì)量為5g．回答并計(jì)算：
（1）求此NaOH溶液中溶質(zhì)的質(zhì)量．
（2）求所得溶液中溶質(zhì)的質(zhì)量分?jǐn)?shù)（酚酞試液質(zhì)量不計(jì)）．

Answer 1

解：設(shè)氫氧化鈉溶液中溶質(zhì)的質(zhì)量為x，生成氯化鈉的質(zhì)量為y
NaOH+HCl═NaCl+H₂O
40 36.5 58.5
x 5g×7.3% y
==
x=0.4g
y=0.585g
所得溶液的質(zhì)量分?jǐn)?shù)=×100%=3.9%
答：氫氧化鈉溶液中溶質(zhì)的質(zhì)量為0.4g，所得溶液中溶質(zhì)的質(zhì)量分?jǐn)?shù)為3.9%．
分析：（1）根據(jù)鹽酸與氫氧化鈉發(fā)生中和反應(yīng)的化學(xué)方程式，可由完全中和時(shí)消耗HCl的質(zhì)量計(jì)算NaOH溶液中溶質(zhì)的質(zhì)量；
（2）由中和反應(yīng)消耗HCl的質(zhì)量計(jì)算出生成氯化鈉的質(zhì)量，由質(zhì)量守恒定律計(jì)算出中和后溶液的質(zhì)量，利用溶質(zhì)質(zhì)量分?jǐn)?shù)計(jì)算公式計(jì)算所得溶液中溶質(zhì)的質(zhì)量分?jǐn)?shù)．
點(diǎn)評(píng)：根據(jù)質(zhì)量守恒定律，反應(yīng)后所得溶液質(zhì)量等于恰好中和時(shí)稀鹽酸的質(zhì)量與氫氧化鈉溶液的質(zhì)量和．