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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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The rate constant for a certain reaction is k = 4.70×10−3 s−1 . If the initial reactant concentration was 0.700 M, what will the
Lynna [10]

Answer:

Therefore the concentration of the reactant after 4.00 minutes will be 0.686M.

Explanation:

The unit of k is s⁻¹.

The order of the reaction = first order.

First order reaction: A first order reaction is  a reaction in which the rate of reaction depends only the value of the concentration of the reactant.

-\frac{d[A]}{dt} =kt

[A] = the concentration of the reactant at time t

k= rate constant

t= time

Here k= 4.70×10⁻³ s⁻¹

t= 4.00

[A₀] = initial concentration of reactant = 0.700 M

-\frac{d[A]}{dt} =kt

\Rightarrow -\frac{d[A]}{[A]}=kdt

Integrating both sides

\Rightarrow\int -\frac{d[A]}{[A]}=\int kdt

⇒ -ln[A] = kt +c

When t=0 , [A] =[A₀]

-ln[A₀]  = k.0 + c

⇒c= -ln[A₀]  

Therefore

-ln[A] = kt - ln[A₀]

Putting the value of k, [A₀] and t

- ln[A] =4.70×10⁻³×4 -ln (0.70)

⇒-ln[A]=  0.375

⇒[A] = 0.686

Therefore the concentration of the reactant after 4.00 minutes will be 0.686M.

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3 years ago
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Explanation:

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Write the balanced reaction and solubility product expression (KSP) for dissolving silver chromate: Ag2CrO4(s). Include all char
Sonbull [250]

Answer:

2Ag⁺ (aq)  + CrO₄⁻² (aq) ⇄  Ag₂CrO₄ (s) ↓

Ksp = [2s]²  . [s] → 4s³

Explanation:

Ag₂CrO₄ → 2Ag⁺  + CrO₄⁻²

Chromate silver is a ionic salt that can be dissociated. When we have a mixture of both ions, we can produce the salt which is a precipitated.

2Ag⁺ (aq)  + CrO₄⁻² (aq) ⇄  Ag₂CrO₄ (s) ↓ Ksp

That's the expression for the precipitation equilibrium.

To determine the solubility product expression, we work with the Ksp

Ag₂CrO₄ (s)  ⇄ 2Ag⁺ (aq)  + CrO₄⁻² (aq)   Ksp

                          2 s                 s

Look the stoichiometry is 1:2, between the salt and the silver.

Ksp = [2s]²  . [s] → 4s³

 

3 0
3 years ago
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