5.)Part B Calcium sulfate, CaSO4, has a K sp value of 7.10×10 5 . What happens w
ID: 891096 • Letter: 5
Question
5.)Part B
Calcium sulfate, CaSO4, has a Ksp value of 7.10×105 . What happens when calcium and sulfate solutions are mixed to give 2.00×103M Ca2+ and 3.00×102M SO42?
Calcium sulfate, , has a value of 7.10×105 . What happens when calcium and sulfate solutions are mixed to give 2.00×103 and 3.00×102 ?
6.)Part A
The addition of which of the following substances will not affect the existing equilibrium in a HF solution?
KCl
7.)Part A
Calculate the pH of a solution that is 0.27 M in HF and 0.11 M in NaF.
Express your answer using two decimal places.
A.) A precipitate forms because Q>Ksp. B.) A precipitate forms because Q<Ksp. C.) No precipitate forms because Q>Ksp. D.) No precipitate forms because Q<Ksp.Explanation / Answer
Solution:-
Q5 part B)
Lets calculate the reaction quotient Q for the CaSO4 using the given concentrations and compare with Ksp
Qc = [Ca^2+][SO4^2-]
Qc = 2.00*10^-3] [3.00*10^-2]
Qc = 6.00*10^-5
The value of the Q is less than the value of the Ksp (7.10*10^-5)
Therefore the correct answer is option D
No precipitate forms because Q < Ksp
Q6 part A)
HF is the weak acid
Therefore its equilibrium can be changed by the addition of the acid , base of conjugate base of the HF
The KCl is the salt which do no thave any common ion with HF
Therefore addition of KCl will not affect the equilibrium of the HF
Therefore the answer is option d. KCl
Q7 Part A)
HF and NaF is the weak acid and its conjugate base therefore they form buffer solution
Calculating the pH of the 0.27 M HF and 0.11 M NaF solution using the Henderson equation
pH= pka + log ([base]/[acid])
pH= 3.17 + log [0.11 /0.27]
pH= 3.17+(-0.39)
pH= 2.78
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