Question

5.)Part B

Calcium sulfate, CaSO4, has a Ksp value of 7.10×10⁻⁵ . What happens when calcium and sulfate solutions are mixed to give 2.00×10⁻³M Ca2+ and 3.00×10⁻²M SO42−?

Calcium sulfate, , has a value of 7.10×10⁻⁵ . What happens when calcium and sulfate solutions are mixed to give 2.00×10⁻³ and 3.00×10⁻² ?

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.

6.)Part A

The addition of which of the following substances will not affect the existing equilibrium in a HF solution?

a.)	KF
	b.)HCl
c.)	NaOH
d.)	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.

Answer 1

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