Consider the following reaction, which leads to the formation of SrCrO4, a highly insoluble salt with Ks (SrCrO4) = 4x10-5
Sr(NO3)2 + K2CrO4 → SrCrO4 + 2KNO3
200 ml of 6.0x10-2 M Sr(NO3)2 are mixed with 100 ml of 0.015 M K2CrO4. If we assume the volume of the mixture is 300 ml,
a)What is the concentration of Sr2+ in the mixture?
b)What is the concentration of CrO4- in the mixture?
c)Will a precipitate of SrCrO4 appear in these conditions
ks of SrCrO4 question, hard 20 point
2006-11-03 6:11 pm
回答 (2)
2006-11-04 4:45 am
✔ 最佳答案
After mixing, the concentration of Sr(NO3)2 = 0.2x6.0x10-2/0.3= 0.04M
After mixing, the concentration of K2CrO4 = 0.1x0.015
= 5x10-3M
Sr(NO3)2 + K2CrO4 → SrCrO4 + 2KNO3
At start: 0.04M 5x10-3M 0M 0M
At eqm:(0.04-x)M (5x10-3-x)M xM 2xM
Ks = [KNO3]^2/[Sr(NO3)2][K2CrO4]
4x10-5 = (2x)^2/(0.04-x)(5x10-3 -x)
solving equqtion, x=4.45x10-5
a)[Sr2+] = 0.04-x
= 0.04-4.45x10-5
= 0.0399555M
b)[CrO4-] = 5x10-3 -x
= 5x10-3 -4.45x10-5
= 4.9555x10-3M
c)No, because the [Sr2+] and [CrO4-] nearly unchange .
2006-11-04 4:35 am
no. of mole of Sr2+ just after mixing = (200/1000)(0.06) = 0.012
no. of mole of CrO4 2- just after mixing = (100/1000)(0.015) = 0.0015
Conc. of Sr2+ just after mixing = 0.012/(300/1000) = 0.04
Conc. of CrO4 2- just after mixing = 0.0015/(300/1000) = 0.005
[Sr2+][CrO4 2-] = 0.04 x 0.005 = 0.0002 > 4X10-5
Therefore, ppt of SrCrO4 will be formed.
2006-11-03 20:35:58 補充：
[Sr2+][CrO4 2-] = 0.04 x 0.005 = 0.0002 greater than 4X10-5
no. of mole of CrO4 2- just after mixing = (100/1000)(0.015) = 0.0015
Conc. of Sr2+ just after mixing = 0.012/(300/1000) = 0.04
Conc. of CrO4 2- just after mixing = 0.0015/(300/1000) = 0.005
[Sr2+][CrO4 2-] = 0.04 x 0.005 = 0.0002 > 4X10-5
Therefore, ppt of SrCrO4 will be formed.
2006-11-03 20:35:58 補充：
[Sr2+][CrO4 2-] = 0.04 x 0.005 = 0.0002 greater than 4X10-5
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