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3 years ago

Question 12 (4 points)

Chemistry

1 answer:

kozerog [31]3 years ago

6 0

Answer:

100grams

Explanation:

law of conservation of mass

if the chemical or whatever, quantity is given then it stay the same in the beginning and the end. Just add them in the end and if something is missing that means you have done something wrong i am just telling for experiment

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3 years ago

How many liters of 0.615 M NaOH will be needed to raise the pH of 0.385 L of 5.13 M sulfurous acid (H2SO3) to a pH of 6.247?

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Answer:

Volume of NaOH required = 3.61 L

Explanation:

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$H2SO3\leftrightarrow H^{+}+HSO3^{-}$ --------(1)

where, Ka1 = 1.5 x 10–2 or pKa1 = 1.824

$HSO3^{-}\leftrightarrow H^{+}+SO3^{2-}$ --------(2)

where, Ka2 = 1.0 x 10–7 or pKa2 = 7.000

The required pH = 6.247 which is beyond the first equivalence point but within the second equivalence point.

Step 1:

Based on equation(1), at the first eq point:

moles of H2SO3 = moles of NaOH

$i.e. \ 5.13\ moles/L*0.385L = moles\ NaOH\\therefore, \ moles\ NaOH = 1.98\ moles\\V(NaOH)\ required = \frac{1.98\ moles}{0.615\ moles/L} =3.22L$

Step 2:

For the second equivalence point setup an ICE table:

$HSO3^{-}+OH^{-}\leftrightarrow H2O+SO3^{2-}$

Initial 1.98 ? 0

Change -x -x x

Equil 1.98-x ?-x x

Here, ?-x =0 i.e. amount of OH- = x

Based on the Henderson buffer equation:

$pH = pKa2 + log\frac{[SO3]^{2-} }{[HSO3]^{-} } \\6.247 = 7.00 + log\frac{x}{(1.98-x)} \\x=0.634 moles$

Volume of NaOH required is:

$\frac{0.634\ moles}{0.615 moles/L}=0.389L$

Step 3:

Total volume of NaOH required = 3.22+0.389 =3.61 L

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