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mezya [45]
3 years ago
8

A 1.0L buffer solution contains 0.100 molHC2H3O2 and 0.100 molNaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5. Calculate the p

H of the solution upon the addition of 10.0 mL of 1.00 MHCl to the original buffer.
Express the your answer to two decimal places.
***will mark Brainliest for right answer ***
Chemistry
1 answer:
Crazy boy [7]3 years ago
7 0
Starting pH = pKa because the acid and conjugate base have equal concentrations.
pH = pKa = -log(1.8*10–5) = 4.74

0.0100 L • 1.00 M = 0.0100 mol HCl, which reacts with the conjugate base, producing 0.0100 mol of weak acid.

You now have 0.110 mol weak acid and 0.090 mol conjugate base. Use the Henderson-Hasselbach equation to solve for pH now.

pH = 4.74 + log(0.090/0.110) = 4.65

The equation technically says to use concentrations of the acid and base, but that doesn’t actually matter, because it would give the same answer.
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A certain liquid X has a normal boiling point of 108.30 °C and a boiling point elevation constant Kb=1.07 °C kg/mol. A solution
Fynjy0 [20]

Answer:

34,6g of (NH₄)₂SO₄

Explanation:

The boiling-point elevation describes the phenomenon in which the boiling point of a liquid increases with the addition of a compound. The formula is:

ΔT = kb×m

Where ΔT is Tsolution - T solvent; kb is ebullioscopic constant and m is molality of ions in solution.

For the problem:

ΔT = 109,7°C-108,3°C = 1,4°C

kb = 1.07 °C kg/mol

Solving:

m = 1,31 mol/kg

As mass of X = 600g = 0,600kg:

1,31mol/kg×0,600kg = 0,785 moles of ions. As (NH₄)₂SO₄ has three ions:

0,785 moles of ions×\frac{1(NH_{4})_{2}SO_{4}}{3Ions} = 0,262 moles of (NH₄)₂SO₄

As molar mass of (NH₄)₂SO₄ is 132,14g/mol:

0,262 moles of (NH₄)₂SO₄×\frac{132,14g}{1mol} = <em>34,6g of (NH₄)₂SO₄</em>

<em></em>

I hope it helps!

7 0
3 years ago
PLEASE HELP!!!!
coldgirl [10]

Answer:

1.2 liters.

Explanation:

Focus on the 4th digit: that's the ones column. The 3rd digit is the decimal place, just be sure to round up.

7 0
3 years ago
In part 3 of the lab, you discovered the relationship between temperature (T) and volume (V) of a gas.
rewona [7]

\frac{V1}{T1} = k

<u>Explanation:</u>

The relation between volume, V of gas and Temperature, T of a gas is related by Charles Law.

This law states that the volume of a given amount of gas held at a constant pressure is directly proportional to the Kelvin temperature

Thus,

\frac{V}{T} = k

where k is a constant

Therefore,

\frac{V1}{T1} = \frac{V2}{T2\\} = \frac{V3}{T3} ...

This shows, as the volume of a gas goes up, the temperature also goes up and vice-versa.

7 0
3 years ago
Please help me!! thank you so much!!​
Elis [28]

Answer:

13.75

Explanation:

gets halved every 8 days so halved twice

4 0
3 years ago
Which element increases its oxidation number in this reaction? 2Na + 2H2O → 2NaOH + H2
sammy [17]

Answer:

Sodium

Explanation:

-Gradpoint

6 0
3 years ago
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