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Natalija [7]
2 years ago
6

A buffer solution contains 0.479 M NaHCO3 and 0.342 M Na2CO3. Determine the pH change when 0.091 mol HNO3 is added to 1.00 L of

the buffer. pH change
Chemistry
1 answer:
pshichka [43]2 years ago
3 0

Answer:

ΔpH = 0.20

Explanation:

The buffer of HCO₃⁻ + CO₃²⁻ has a pka of 10.2

HCO₃⁻ ⇄ H⁺ + CO₃²⁻

There are 0.479moles of NaHCO₃ and 0.342moles of Na₂CO₃.

Using Henderson-Hasselbalch formula:

pH = pka + log [Base] / [Acid]

pH = 10.2 + log 0.342mol / 0.479mol

<em>pH = 10.05</em>

NaOH reacts with HCO₃⁻ producing CO₃²⁻, thus:

NaOH + HCO₃⁻ → CO₃²⁻ + H₂O + Na⁺

0.091 moles of NaOH produce the same moles of CO₃²⁻ and consume HCO₃⁻. Moles of these species are:

CO₃²⁻: 0.342mol + 0.091mol: 0.433mol

HCO₃⁻: 0.479mol - 0.091 mol: 0.388mol

Using Henderson-Hasselbalch formula:

pH = pka + log [Base] / [Acid]

pH = 10.2 + log 0.433mol / 0.388mol

pH = 10.25

That means change of pH, ΔpH is:

ΔpH = 10.25 - 10.05 = <em>0.20</em>

<em />

I hope it helps!

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<u>Answer:</u> The value of K_p for the reaction is 6.32 and concentrations of (CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl is 0.094 M, 0.094 M and 0.106 M respectively.

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\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}

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So, \text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M

For the given chemical reaction:

                (CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)

Initial:               0.2                    -                        -

At Eqllm:          0.2 - x               x                       x

The expression of K_c for above reaction follows:

K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}

Putting values in above equation, we get:

0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

[(CH_3)_2C=CH]=x=0.094M

[HCl]=x=0.094M

[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M

Hence, the value of K_p for the reaction is 6.32 and concentrations of (CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl is 0.094 M, 0.094 M and 0.106 M respectively.

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