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3 years ago

Enter your answer in the provided box.

Chemistry

1 answer:

nexus9112 [7]3 years ago

4 0

Answer:

V₂ = 50.93 L

Explanation:

Initial volume, $V_1=43.1\ L$

Initial temperature, $T_1=24^{\circ} C=24+273=297\ K$

Final temperature, $T_2=78^{\circ} C=78+273=351\ K$

We need to find the final volume of the gas. The relation between the volume and the temperature is given by :

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{43.1\times 351}{297}\\\\V_2=50.93\ L$

So, the final volume of the gas is 50.93 L.

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A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th

Mashcka [7]

Explanation:

It is known that $K_{a}$ of $HNO_{2}$ = $4.5 \times 10^{-4}$ .

(a) Relation between $K_{a}$ and $pK_{a}$ is as follows.

$pK_{a} = -log (K_{a})$

Putting the values into the above formula as follows.

$pK_{a} = -log (K_{a})$

= $-log(4.5 \times 10^{-4})$

= 3.347

Also, relation between pH and $pK_{a}$ is as follows.

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= 0.0116 mol

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Hence, before the reaction:

No. of moles of $NO^{-}_{2}$ = $0.15 M \times 1.0 L$

= 0.15 mol

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= 0.12 mol

On the other hand, after the reaction :

No. of moles of $NO^{-}_{2}$ = moles present initially - moles added

= (0.15 - 0.0116) mol

= 0.1384 mol

Moles of $HNO_{2}$ = moles present initially + moles added

= (0.12 + 0.0116) mol

= 0.1316 mol

As, $K_{a}$ = $4.5 \times 10^{-4}$

$pK_{a} = -log (K_{a})$

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Since, volume is both in numerator and denominator, we can use mol instead of concentration.

As, pH = $pK_{a} + log \frac{[conjugate base]}{[acid]}$

= 3.347+ log {0.1384/0.1316}

= 3.369

= 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

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Answer:

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