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nignag [31]
3 years ago
15

A buffer contains 0.18 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. What is the pH

of this buffer?
What is the pH of the buffer after the addition of 0.02 mol of NaOH?

What is the pH of the buffer after the addition of 0.02 mol of HI?
Chemistry
1 answer:
Yuki888 [10]3 years ago
8 0

Answer:

1) pH = 5.05

2) pH = 5.13

3) pH = 4.97

Explanation:

Step 1: Data given

Number of moles of propionic acid = 0.18 moles

Number of moles sodium propionate = 0.26 moles

Volume = 1.20 L

Ka = 1.3 * 10^-5    → pKa = 4.989

Step 2: Calculate concentrations

Concentration = moles / volume

[acid]= 0.18/ 1.2 =0.150 M

[salt]= 0.26/ 1.3 = 0.217 M

pH = 4.89 + log(0.217/0.150)=<u>5.05</u>

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What is the pH of the buffer after the addition of 0.02 mol of NaOH?

moles acid = 0.18 - 0.02 = 0.16

[acid]= 0.16/ 1.2=0.133 M

moles salt = 0.26 + 0.02 = 0.28

[salt]= 0.28/ 12=0.233

pH = 4.89 + log 0.233/ 0.133 = 5.13

What is the pH of the buffer after the addition of 0.02 mol of HI?

moles acid = 0.18+ 0.02 = 0.20 moles

[acid]= 0.20/ 1.2 = 0.167 M

[salt]= 0.26 - 0.02= 0.24 moles

[salt]= 0.24/ 1.2 = 0.20 M

pH = 4.89 + log 0.20/ 0.167= 4.97

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How many times greater is the hydronium ion concentration in rainfall (pH = 5.0) compared to pure water (pH = 7.0)?
joja [24]
Hydronium ion is H3O(+)

pH is log 1 / [H3O(+)] = - log [H3O(+)]

Rainfall: pH = 5.0 = - log [H3O(+)] => [H3O(+)] = -antilog (- 5.0) =  10^-5

Water; pH = 7.0 = - log [H3O(+)] => [H3O(+)] = - antilog (-7.0) = 10^ -7

Divide hidronium conentration of rainfall by hidrodium concentration of water:

10^-5 / 10^-7 = 10^2 = 100.

Then, hidronium concentration in rainfall is 100 times hidronium concentration in water.

Answer: option a. 100 times.
3 0
3 years ago
For each image, select the state of matter represented.
kifflom [539]

Answer:

need image

Explanation:

5 0
3 years ago
A certain liquid X has a normal freezing point of 7.60 °C and a freezing point depression constant K= 6.90 °C-kg-mol. Calculate
Dmitry [639]

<u>Answer:</u> The freezing point of solution is -5.11°C

<u>Explanation:</u>

Vant hoff factor for ionic solute is the number of ions that are present in a solution. The equation for the ionization of sodium chloride follows:

NaCl(aq.)\rightarrow Na^{+}(aq.)+Cl^-(aq.)

The total number of ions present in the solution are 2.

To calculate the molality of solution, we use the equation:

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

Where,

m_{solute} = Given mass of solute (NaCl) = 7.57 g

M_{solute} = Molar mass of solute (NaCl) = 58.44 g/mol

W_{solvent} = Mass of solvent (liquid X) = 350.0 g

Putting values in above equation, we get:

\text{Molality of }NaCl=\frac{7.57\times 1000}{58.44\times 350.0}\\\\\text{Molality of }NaCl=0.370m

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

\Delta T=iK_fm

where,

i = Vant hoff factor = 2

K_f = molal freezing point depression constant = 6.90°C/m

m = molality of solution = 0.370 m

Putting values in above equation, we get:

\Delta T=2\times 6.90^oC/m.g\times 0.370m\\\\\Delta T=5.11^oC

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

\Delta T=\text{freezing point of water}-\text{freezing point of solution}

\Delta T = 5.11 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

5.106^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-5.11^oC

Hence, the freezing point of solution is -5.11°C

7 0
4 years ago
How is the Periodic Table helpful to scientists?
s2008m [1.1K]

Answer:

Because it helps predict the types of chemical reactions that a particular element is likely to participate in by telling each element's atomic number and usually its atomic weight

Explanation:

5 0
3 years ago
Read 2 more answers
What is the (OH-) in a solution with a pOH of 6.48
inn [45]
Through manipulation of equations, we are able to obtain the equation:

-pOH= log [ OH^{-}]

Then we can transform the equation into:

[ OH^{-}]= 10^{-pOH}

Then we are able to plug in the pOH and directly get [OH-]:

[ OH^{-}] = 10^{-6.48}

[ OH^{-}]=3.31* 10^{-7} M
3 0
3 years ago
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