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3 years ago

Calculate the molarity of a solution consisting of 65.5 g of K2S0 4 in 5.00 L of solution.

Chemistry

1 answer:

tia_tia [17]3 years ago

8 0

Answer:

Molarity is 0.075 M. 

Explanation:

Moles:

${ \tt{ = \frac{65.5}{RFM} }}$

RFM of potassium sulphate :

${ \tt{ = (39 \times 2) + 32 + (16 \times 4)}} \\ = 174 \: g$

substitute:

${ \tt{moles = \frac{65.5}{174} = 0.376 \: moles}}$

In volume of 5.00 l:

${ \tt{5.00 \: l = 0.376 \: moles}} \\ { \tt{1 \: l = ( \frac{0.376}{5.00} ) \: moles}} \\ { \tt{molarity = 0.075 \: mol \: l {}^{ - 1} }}$

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a) We have to add 0.9 mole NaOH to 1.0 L of 1.8 M HC2H3O2

b) We have to add 0.277 mole NaOH to 1.0 L of 1.8 M HC2H3O2

c) We have to add 1.16 mole NaOH to 1.0 L of 1.8 M HC2H3O2

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a) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = pKa?

Step 1: Data given

Volume of HC2H3O2 = 1.0 L

Molarity of HC2H3O2 = 1.8 M

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Use the Henderson-Hasselbalch equation.

pH = pKa + log(A-/HA)

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Remaining moles of HA = 1.8 - X

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HA = 1.8 - X

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We have to add 0.9 mole NaOH to 1.0 L of 1.8 M HC2H3O2

To control we can do the following equation:

4.74 = 4.74 + log(0.9/0.9) = 4.74

b) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = 4.00?

Step 1: Data given

Volume of HC2H3O2 = 1.0 L

Molarity of HC2H3O2 = 1.8 M

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Use the Henderson-Hasselbalch equation.

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A-/HA = 0.182

Consider X = moles of NaOH added (and moles of A- formed)

Remaining moles of HA = 1.8 - X

moles of A- = X

HA = 1.8 - X

X/(1.8-X) = 0.182

X =0.277

We have to add 0.277 mole NaOH to 1.0 L of 1.8 M HC2H3O2

To control we can do the following equation:

4 = 4.74 + log(0.277/1.523)

c) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = 5.00

Step 1: Data given

Volume of HC2H3O2 = 1.0 L

Molarity of HC2H3O2 = 1.8 M

Ka = 1.8*10^-5

ph = 5

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Use the Henderson-Hasselbalch equation.

pH = pKa + log(A-/HA)

5 = 4.74 + log(A-/HA)

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Remaining moles of HA = 1.8 - X

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HA = 1.8 - X

X/(1.8-X) = 1.82

X =1.16

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Calculate the molarity of a solution consisting of 65.5 g of K2S0 4 in 5.00 L of solution. ​

Calculate the molarity of a solution consisting of 65.5 g of K2S0 4 in 5.00 L of solution.