All categories

3 years ago

Consider the reversible dissolution of lead(II) chloride.

Chemistry

1 answer:

IrinaVladis [17]3 years ago

5 0

Answer:

8.96x10^-4

Explanation:

Step 1:

The equation for the reaction is given below:

PbCl2(s) <==> Pb^2(aq) + + 2Cl^-(aq)

Step 2:

Data obtained from the question. This includes:

Mass of PbCl2 = 0.2495 g

Volume of solution = 50.0 mL

Concentration of Pb^2+, [Pb^2] = 0.0159 M

Concentration of Cl^-, [Cl^-] = 0.0318 M

Equilibrium constant Kc =?

Step 3:

Determination of the mole of PbCl2 in 0.2495 g of PbCl2. This is illustrated below:

Mass of PbCl2 = 0.2495 g

Molar Mass of PbCl2 = 207 + (35.5x2) = 207 + 71 = 278g/mol

Number of mole = Mass/Molar Mass

Number of mole PbCl2 = 0.2495/278

Number of mole PbCl2 = 8.97x10^-4 mole

Step 4:

Determination of the molarity of PbCl2. This is shown below:

Mole of PbCl2 = 8.97x10^-4 mole

Volume = 50 mL = 50/1000 = 0.05 L

Molarity of PbCl2 =?

Molarity = mole /Volume

Molarity of PbCl2 = 8.97x10^-4/0.05

Molarity of PbCl2 = 0.01794 M

Step 5:

Determination of the equilibrium constant Kc. This is illustrated below:

PbCl2(s) <==> Pb^2(aq) + + 2Cl^-(aq)

The equilibrium constant for the reaction above is given by:

Kc = [Pb^2] [Cl^-]^2 / [PbCl2]

[Pb^2] = 0.0159 M

[Cl^-] = 0.0318 M

[PbCl2] = 0.01794 M

Kc =?

Kc = [Pb^2] [Cl^-]^2 / [PbCl2]

Kc = 0.0159 x (0.0318)^2 /0.01794

Kc = 8.96x10^-4

You might be interested in

Consider an exceptionally weak acid, HA, with a Ka = 1x10 -20 . You make a 0.1M solution of the salt Na

Vedmedyk [2.9K]

Answer:

$pH=13$

Explanation:

Hello,

In this case, given the acid, we can suppose a simple dissociation as:

$HA\rightleftharpoons H^+ + A^-$

Which occurs in aqueous phase, therefore, the law of mass action is written by:

$Ka=\frac{[H^+][A^-]}{[HA]}$

That in terms of the change $x$ due to the reaction's extent we can write:

$1x10^{-20}=\frac{x*x}{0.1M-x}$

But we prefer to compute the Kb due to its exceptional weakness:

$Kb=\frac{Kw}{Ka}=\frac{1x10^{-14}}{1x10^{-20}} =1x10^{-6}$

Next, the acid dissociation in the presence of the base we have:

$Kb=\frac{[OH^-][HA]}{[A^-]}=1x10^{6}=\frac{x*x}{0.1-x}$

Whose solution is $x=0.0999M$ which equals the concentration of hydroxyl in the solution, thus we compute the pOH:

$pOH=-log([OH^-])=-log(0.0999)=1$

Finally, since the maximum scale is 14, we can compute the pH by knowing the pOH:

$pH+pOH=14\\\\pH=14-pOH=14-1\\\\pH=13$

Regards.

5 0

3 years ago

Read 2 more answers

According to Le Châtelier's principle, a system at equilibrium will respond to a stress by shifting in the direction that reliev

Romashka [77]

Answer:

a. reduces the stress

Explanation:

3 0

3 years ago

Which of the following particles combine to form molecules?

solniwko [45]

Answer: i think its protons and electrons but it also might just be atoms because protons and electrons make atoms when there are also neutrons

Explanation:

7 0

3 years ago

Five students performed a Kjeldahl nitrogen analysis of a protein sample. The following weight % nitrogen values were determined

Sunny_sXe [5.5K]

Answer:

G_calculated = 1.756

The outlier should be rejected, as G_cal > G_tab (= 1.463) at 95 % confidence.

Explanation:

The Grubb's test is used for identifying an outlier in data, which is from the same population. For this, a statistical term, G, is calculated for the suspected outlier. If the calculated value is greater than the tabulated G value then the suspected value is rejected. This term is given as,

G_calculated = | suspect value - mean| / s

Here, suspect value is 13.8, mean is to be taken of all the data (including suspected value). s is the standard deviation of the sample data.

s is calculated from the following formula:

s = (Σ(xi - x)²/(N-1))^1/2

Here, x is the mean, which is 15.24, xi is individual value and N is the total number of data (5).

From the above formula, s is found to be

Standard Deviation, s = 0.820

Now for G value,

G_calculated = | 13.8 - 15.24| / (0.820)

G_ calculated = 1.756

The tabulated G value at 95 % confidence and N -1 (5 - 1 = 4) degree of freedom is, 1.463.

As calculated G (1.756) is greater than the tabulated G (1.463), the value 13.8 is considered an outlier at 95 % confidence.

3 0

3 years ago

How many grams of carbon are contained in 2.25 g of potassium carbonate, K2CO3

romanna [79]

The mass of carbon contained in 2.25 g of potassium carbonate, K₂CO₃ is 0.196 g.

<h3>Molecular mass of potassium carbonate</h3>

The molecular mass of potassium carbonate, K₂CO₃ is calculated as follows;

M = K₂CO₃

M = (39 x 2) + (12) + (16 x 3)

M = 138 g

mass of carbon in potassium carbonate, K₂CO₃ is = 12 g

The mass of carbon contained in 2.25 g of potassium carbonate, K₂CO₃ is calculated as follows;

138 g ------------ 12 g of carbon

2.25 g ------------ ?

= (2.25 x 12) / 138

= 0.196 g

Thus, the mass of carbon contained in 2.25 g of potassium carbonate, K₂CO₃ is 0.196 g.

Learn more about potassium carbonate here: brainly.com/question/27514966

#SPJ1

5 0

1 year ago