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3 years ago

How would you prepare 500 ml of the following solutions : Sodium succinate buffer (0.1 mol/dm3 pH 5.64)

Chemistry

1 answer:

Nana76 [90]3 years ago

7 0

Answer:

8.10g of sodium succinate must be added and 247mL of 0.1M HCl adding enough water until make 500mL

Explanation:

<em>Succinic acid has a pKa₂ of 5.63</em>

To solve this question we must find the amount of sodium succinate and 0.1M HCl that we have to add using H-H equation:

pH = pKa + log [A-] / [HA]

5.64 = 5.63 + log [Na₂Succ.] / [HSucc⁺]

0.01 = log [Na₂Succ.] / [HSucc⁺]

1.0233 = [Na₂Succ.] / [HSucc⁺] <em>(1)</em>

As:

0.1M = [Na₂Succ.] + [HSucc⁺] <em>(2)</em>

Replacing (2) in (1):

1.0233 = 0.1M - [HSucc⁺] / [HSucc⁺]

1.0233[HSucc⁺] = 0.1M - [HSucc⁺]

2.033[HSucc⁺] = 0.1M

[HSucc⁺] = 0.0494M

[Na₂Succ] = 0.0506M

Both [Na₂Succ⁺] and [HSucc⁺] ions comes from the same sodium succinate we have to find the moles of sodium succinate in 500mL of 0.1M. Then, based on the reaction:

Na₂Succ + HCl → HSucc⁺ + Cl⁻

The moles of HCl added = Moles HSucc⁺ we need:

<em>Moles Na₂Succ:</em>

0.500L * (0.1mol/L) = 0.0500 moles

<em>Mass -Molar mass sodium succinate: 162.05g/mol-:</em>

0.0500mol * (162.05g/mol) = 8.10g of sodium succinate must be added

<em>Moles HCl:</em>

0.0494M * 0.500L = 0.0247 moles HCl * (1L / 0.1mol) = 0.247L =

And 247mL of 0.1M HCl adding enough water until make 500mL

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1 year ago

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3 years ago

Read 2 more answers

A glucose solution that is prepared for a patient should have a concentration of 180 g/L. A nurse has 18 g of glucose. How

Irina-Kira [14]

<h3>Answer:</h3>

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<h3>Explanation:</h3>

The concentration of glucose is given as 180 g/L

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Concentration in g/L is calculated by dividing mass of the solute by the volume of the solution.
When calculating molarity on the other hand, we divide number of moles of the solute by the volume of the solution.

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Rearranging the formula,

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