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Serggg [28]
3 years ago
7

A buffer solution contains 0.11 mol of acetic acid and 0.13 mol of sodium acetate in 1.00 L. What is the pH of this buffer?

Chemistry
1 answer:
aleksley [76]3 years ago
5 0

Answer:

Henderson Hasselbalch equation: pH = pKa + log [salt]/[acid]

You need to know the pKa for acetic acid. Looking it up one finds it to be 4.76

(a). pH = 4.76 + log [0.13]/[0.10]

= 4.76 + 0.11

= 4.87

(b) KOH + CH3COOH =>H2O + CH3COOK so (acid)goes down and (salt)goes up. Assuming no change in volume, you have 0.10 mol acid - 0.02 mol = 0.08 mol acid and 0.13 mol salt + 0.02 mol = 0.15 mol salt

pH = 4.76 + log [0.15]/[0.08]

= 4.76 + 0.27

= 5.03

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At 298 K, the osmotic pressure of a glucose solution (C6H12O6 (aq)) is 12.1 atm. Calculate the freezing point of the solution. T
Anarel [89]

<u>Answer:</u> The freezing point of solution is -0.974°C

<u>Explanation:</u>

  • To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

where,

\pi = osmotic pressure of the solution = 12.1 atm

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = ?

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the solution = 298 K

Putting values in above equation, we get:

12.1atm=1\times M\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\M=\frac{12.1}{1\times 0.0821\times 298}=0.495M

This means that 0.495 moles of glucose is present in 1 L or 1000 mL of solution

  • To calculate the mass of solution, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of solution = 1.034 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

1.034g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.034g/mL\times 1000mL)=1034g

  • To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of glucose = 0.495 moles

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

0.495mol=\frac{\text{Mass of glucose}}{180.16g/mol}\\\\\text{Mass of glucose}=(0.495mol\times 180.16g/mol)=89.18g

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

  • The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point elevation constant = 1.86°C/m

m_{solute} = Given mass of solute (glucose) = 89.18 g

M_{solute} = Molar mass of solute (glucose) = 180.16  g/mol

W_{solvent} = Mass of solvent (water) = [1034 - 89.18] g = 944.82 g

Putting values in above equation, we get:

0-\text{Freezing point of solution}=1\times 1.86^oC/m\times \frac{89.18\times 1000}{180.16g/mol\times 944.82}\\\\\text{Freezing point of solution}=-0.974^oC

Hence, the freezing point of solution is -0.974°C

8 0
3 years ago
Ammonia (NH3) reacts with sulfuric acid to form ammonium sulfate. How many grams of ammonium sulfate are obtained from 8.73 g of
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I hope you understand my working:
1) Finding the mol of NH3 to find the mol of (NH4)2SO4 (ammonium sulfate)
2) Mr of (NH4)2SO4
3) Theoretical yield: The actual grams of (NH4)2SO4 produced when reacting 0.514 mol of NH3 to 0.514 mol H2SO4
4) Using formula of (given grams)/(theoretical grams or actual grams) * 100 = 73%
5) Basic algebra

4 0
3 years ago
I don’t understand how to do this
djverab [1.8K]
Carbon dioxide is a carbon atom and two oxygen atoms, therefore CO2

For the others, they are hydrocarbons.
The first part of the name is determined by how many carbon atoms there are. The second part, is by the type. Alcohol, Alkane, Alkene, alkynes, acid, esters, amides.

7 0
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The figure above shows the Earth at two different positions in its orbit around the Sun. Which position corresponds to summer in
Lelu [443]

Answer:

A. Position B.

Explanation:

In the summer the northern hemisphere is closer to the Sun so it's Position B.

3 0
3 years ago
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If there are 40 mol of NBr3 and 48 mol of NaOH, what is the excess reactant?
Nata [24]

Answer:

The correct answer is option B.

Explanation:

3NaOH+2NBr_3\rightarrow 3HOBr+3NaBr+N_2

Moles of NBr_3 = 40 mol

Moles of NaOH = 48 mol

According to reaction, 3 moles of NaOH reacts with 2 moles NBr_3

Then ,48 moles of NaOH will reacts with:

\frac{2}{3}\times 48 mol=32 mol of NBr_3

Then ,40 moles of NaBr_3 will reacts with:

\frac{3}{2}\times 40 mol=60 mol of NaOH

As we can see that 48 moles of sodium will completey react with 32 moles of nitrogen tribromide.

Moles left after reaction = 40 mol - 32 mol = 8 mol

Hence, the NBr_3 is an excessive reagent.

6 0
3 years ago
Read 2 more answers
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