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jeyben [28]
3 years ago
10

A concentrated aqueous solution of Pb(NO3)2 is slowly added to 1.0 L of a mixed aqueous solution containing 0.010 M Na2CrO4 and

2.5 M NaBr in an attempt to selectively precipitate only one of the anions. The sparingly soluble salt that precipitates first (neglecting any volume change) is ________ with a [Pb2+] concentration of ________.
Ksp(PbCrO4) = 1.8 × 10–14, Ksp(PbBr2) = 6.3 × 10–6
Chemistry
1 answer:
Oduvanchick [21]3 years ago
5 0

Answer:

This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M

Explanation:

Step 1: Data given

Molarity of Na2CrO4 = 0.010 M

Molarity of NaBr = 2.5 M

Ksp(PbCrO4) = 1.8 * 10^–14

Ksp(PbBr2) = 6.3 * 10^–6

Step 2: The balanced equation

PbCrO4 →Pb^2+ + CrO4^2-

PbBr2  → Pb^2+ + 2Br-

Step 3: Define Ksp

Ksp PbCrO4 = [Pb^2+]*[CrO4^2-]

1.8*10^-14 = [Pb^2+] * 0.010 M

[Pb^2+] = 1.8*10^-14 /0.010

[Pb^2+] = 1.8*10^-12 M

The minimum [Pb^2+] needed to precipitate PbCrO4 is 1.8*10^-12 M

Ksp PbBr2 = [Pb^2+][Br-]²

6.3 * 10^–6 = [Pb^2+] (2.5)²

[Pb^2+] = 1*10^-6 M

The minimum [Pb^2+] needed to precipitate PbBr2 is 1*10^-6 M

This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M

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Determine the pHpH of an HFHF solution of each of the following concentrations. In which cases can you not make the simplifying
PIT_PIT [208]

The question is incomplete, complete question is :

Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? (K_a for HF is 6.8\times 10^{-4}.)

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Express your answer to two decimal places.

Answer:

The pH of an 0.280 M HF solution is 1.87.

Explanation:3

Initial concentration if HF = c = 0.280 M

Dissociation constant of the HF = K_a=6.8\times 10^{-4}

HF\rightleftharpoons H^++F^-

Initially

c          0            0

At equilibrium :

(c-x)      x             x

The expression of disassociation constant is given as:

K_a=\frac{[H^+][F^-]}{[HF]}

K_a=\frac{x\times x}{(c-x)}

6.8\times 10^{-4}=\frac{x^2}{(0.280 M-x)}

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x = 0.01346 M

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The pH of the solution is ;

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