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Fynjy0 [20]
2 years ago
11

What is the [ch3co2-]/[ch3co2h] ratio necessary to make a buffer solution with a ph of 4.34? ka = 1.8 × 10-5 for ch3co2h?

Chemistry
1 answer:
xeze [42]2 years ago
5 0
According to Henderson–Hasselbalch Equation,

                                    pH  =  pKa + log [Acetate] / [Acetic Acid]

As,
           pKa = -log Ka
           pKa = -log (1.8 × 10⁻⁵)
           pKa =  4.74
So,
                               pH  =  4.74 + log [Acetate] / [Acetic Acid]

                                  4.34  =  4.74 + log [Acetate] / [Acetic Acid]

                        4.34 - 4.74  = log [Acetate] / [Acetic Acid]

                                 -0.40  =  log [Acetate] / [Acetic Acid]

Taking Antilog on both sides,

               [Acetate] / [Acetic Acid]  =  0.398
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3 years ago
Please help me and explanation would be really awesome thank you!
valentina_108 [34]

Answer:

So the answer would be 10 moles

Explanation:

1) Start with the molecular formula for water: H_{2} O!

2) If there are 10 moles of water use a mole ratio to calculate the moles of oxygen it would produce.

(This question is... interesting... since they chose an element that is diatomic in free state so It could TECHNICALLY be two answers, moles of O or moles of O_{2})

The mole ratio is 1 moles of H_{2}O to 1 moles of O. This is because the coefficient for oxygen in water is simple 1, so the ratio is 1:1.

3) that means if 10 moles of water decompose, they decompose into 10 moles of H_{2} and 10 moles of O.

Extra:

About what I was saying before about the question being slightly interesting:

10 moles of pure oxygen is produced but free state oxygen exists as O_{2} so it could possibly be 10 OR 5! However, notice it says elements. This leads me to believe the answer is 10 (monatomic oxygen) instead of 5 (free state/diatomic oxygen).

I hope this helps!

4 0
3 years ago
If a balloon initially has a volume of 4.0 liters and a temperature of 10 degrees Celsius, what will the volume of the balloon b
Anika [276]
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7 0
2 years ago
Solid sodium iodide is slowly added to a solution that is 0.0050 M Pb 2+ and 0.0050 M Ag +. [K sp (PbI 2) = 1.4 × 10 –8; K sp (A
UkoKoshka [18]

Answer:

[Ag⁺] = 5.0x10⁻¹⁴M

Explanation:

The product solubility constant, Ksp, of the insoluble salts PbI₂ and AgI is defined as follows:

Ksp(PbI₂) = [Pb²⁺] [I⁻]² = 1.4x10⁻⁸

Ksp(AgI) = [Ag⁺] [I⁻] = 8.3x10⁻¹⁷

The PbI₂ <em>just begin to precipitate when the product  [Pb²⁺] [I⁻]² = 1.4x10⁻⁸</em>

<em />

As the initial [Pb²⁺] = 0.0050M:

[Pb²⁺] [I⁻]² = 1.4x10⁻⁸

[0.0050] [I⁻]² = 1.4x10⁻⁸

[I⁻]² = 1.4x10⁻⁸ / 0.0050

[I⁻]² = 2.8x10⁻⁶

<h3>[I⁻] = 1.67x10⁻³</h3><h3 />

So, as the [I⁻] concentration is also in the expression of Ksp of AgI and you know concentration in solution of I⁻ = 1.67x10⁻³M:

[Ag⁺] [I⁻] = 8.3x10⁻¹⁷

[Ag⁺] [1.67x10⁻³] = 8.3x10⁻¹⁷

<h3>[Ag⁺] = 5.0x10⁻¹⁴M</h3>

6 0
3 years ago
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GaryK [48]

Answer:

234

Explanation:

8 0
2 years ago
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