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Schach [20]
2 years ago
12

URGENT ! PLEASE ANSWER QUICKLY

Chemistry
2 answers:
Bumek [7]2 years ago
7 0

Answer:If we dissolve NaF in water, we get the following equilibrium:

text{F}^-(aq)+text{H}_2text{O}(l) rightleftarrows text{HF}(aq)+text{OH}^-(aq)

The pH of the resulting solution can be determined if the  K_b of the fluoride ion is known.

20.0 g of sodium fluoride is dissolve in enough water to make 500.0 mL of solution. Calculate the pH of the solution. The  K_b of the fluoride ion is 1.4 × 10 −11 .

Step 1: List the known values and plan the problem.

Known

mass NaF = 20.0 g

molar mass NaF = 41.99 g/mol

volume solution = 0.500 L

K_b of F – = 1.4 × 10 −11

Unknown

pH of solution = ?

The molarity of the F − solution can be calculated from the mass, molar mass, and solution volume. Since NaF completely dissociates, the molarity of the NaF is equal to the molarity of the F − ion. An ICE Table (below) can be used to calculate the concentration of OH − produced and then the pH of the solution.

Explanation:

marysya [2.9K]2 years ago
4 0

Answer:

not sure if this is 100% correct but this is what i got first question: molarity=mol/L 5g*1mol/58.44g=5/58.44 =0.085558mol 4.8m=0.085558/L L=0.0178

jiujuan avatar

second question: pH = - log [H3O+] pH= -log [4.6*10^-9] = 8.33724216

Explanation:

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Lead thiocyanate, Pb(SCN)2, has a Ksp value of 2.00×10−5.a) Calculate the molar solubility of lead thiocyanate in pure water. Th
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a) 1.71 × 10⁻³ M

b) 8.00 × 10⁻⁵ M

Explanation:

In order to calculate the solubility (S) of Pb(SCN)₂ we will use an ICE chart. We identify 3 stages (Initial, Change, Equilibrium) and complete each row with the concentration or change in the concentration.

       Pb(SCN)₂(s) ⇄ Pb²⁺(aq) + 2 SCN⁻(aq)

I                                 0                   0

C                               +S               +2S

E                                 S                  2S

The solubility product (Ksp) is:

Ksp = 2.00 × 10⁻⁵ = [Pb²⁺].[SCN⁻]² = S . (2S)² = 4S³

S = 1.71 × 10⁻³ M

<em>b) Calculate the molar solubility of lead thiocyanate in 0.500 M KSCN.</em>

KSCN is a strong electrolyte that dissociates to give 0.500 M K⁺ and 0.500M SCN⁻.

       Pb(SCN)₂(s) ⇄ Pb²⁺(aq) + 2 SCN⁻(aq)

I                                 0                 0.500

C                               +S                +2S

E                                 S                0.500 + 2S

Ksp = 2.00 × 10⁻⁵ = [Pb²⁺].[SCN⁻]² = S . (0.500 + 2S)²

In the term (0.500 + 2S)², 2S is negligible.

Ksp = 2.00 × 10⁻⁵ = S . (0.500)²

S = 8.00 × 10⁻⁵ M

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