分析化学有效数字运算规则习题及答案 《分析化学习题答案及详解》高等教育出版社(13)

δ 1 (hco 3 ) =

10 7 .10 × 10 6 .38 = 0.839 (10 7 .10 ) 2 + 10 6 .38 × 10 7 .10 + 10 6 .38 × 10 10 .25

δ 0 (co32 ) =

ph=8.32

10 6.38 × 10 10.25 = 0.001 (10 7.10 ) 2 + 10 6.38 × 10 7.10 + 10 6.38 × 10 10.25

δ 2 ( h 2 co 3 ) =

δ 1 ( hco 3 ) =

(10 8 .32 ) 2 = 0.0112 (10 8 .32 ) 2 + 10 6 .38 × 10 8 .32 + 10 10 .25 × 10 6 .38

10 8 .32 × 10 6 .38 = 0.979 (10 8 .32 ) 2 + 10 6 .38 × 10 8 .32 + 10 6 .38 × 10 10 .25

10 6 .38 × 10 10 .25 = 0.0115 (10 8 .32 ) 2 + 10 6 .38 × 10 8 .32 + 10 6 .38 × 10 10 .25

2 δ 0 ( co 3 ) =

ph=9.50

δ 2 ( h 2 co 3 ) =

δ 1 ( hco 3 ) =

(10 9 .50 ) 2 = 6.34×10-4 (10 9 .50 ) 2 + 10 6 .38 × 10 9 .50 + 10 10 .25 × 10 6 .38

10 9 .50 × 10 6 .38 = 0.851 (10 9 .50 ) 2 + 10 6 .38 × 10 9 .50 + 10 6 .38 × 10 10 .25

10 6 .38 × 10 10 .25 = 0.150 (10 9 .50 ) 2 + 10 6 .38 × 10 9 .50 + 10 6 .38 × 10 10 .25

δ 0 (co

2 3

)=

5．已知 hoac 的 pka =4.74，nh3·h20 的 pkb=4.74。计算下列各溶液的 ph： (1) 0.10 mol·l-1hoac； (2) 0.10 mol·l-1nh3·h2o； (4) 0.15 mol·l-1naoac。 (3) 0.15 mo1·l-1nh4cl； 解：(1) 0.1mol·l-1hac

∵ c / ka =

0 .1 ＞ 10 10 5

c k a = 0 . 1 × 10 4 .74 ＞ 10 k w

4 . 74 = 1 .35 × 10 3 (mol l1 ) ∴〔h+〕= 0 .1 × 10

ph=2.87 (2) 0.10 mol·l-1nh3·h2o

q c / kb = ∴ [ oh ] = ph = 11.13 0 .1 ＞ 105 10 4 .74 c k b = 0 .1 × 10 4 .74 ＞ 10 k w

0 . 1 × 10 4 .74 = 1 . 35 × 10 3 (mol l1 )

(3) 0.15 mo1·l-1nh4cl

q c / k a= 0 . 15 ＞ 105 10 9 .26 c k a = 0 . 15 × 10 9 .26 ＞ 10 k w 0 . 15 × 10 9 .26 = 9 . 03 × 10 6 (mol l 1 ) ph = 5 . 04 (4) 0.15 mol·l-1naoac