The next three questions pertain to the preparation of a buffer starting with th

Question

The next three questions pertain to the preparation of a buffer starting with the salt of the weak base and adding an amount of strong acid to convert some salt to its weak acid. The amount of strong acid is selected to achieve the desired amount of the conjugate pair concentration ratio at the specified pH.

3. 4. 5.

You need to prepare 100 mL of a solution that is 0.05 M sodium bicarbonate. Calculate amount sodium bicarbonate you need to weigh out to prepare this solution.

This part I was able to answer and got 0.42g needed. I need help with the two below. Thanks :)

You want to convert a portion of the bicarbonate in this solution to carbonic acid with a molarity of 0.02M utilizing 1 M HCl. How much HCl do you add? (Ignore the slight change in volume.)

What will be the molarities of the two components in the solution?

Explanation / Answer

You need to prepare 100 mL of a solution that is 0.05 M sodium bicarbonate

Moles of sodium bicarbonate = Molarity * Volume ( in L) = 0.05 M * (100 / 1000) L = 0.005 mol

Mass of sodium bicarbonate = Moles * molar mass = 0.005 mol * 84 g/mol = 0.42 g

You want to convert a portion of the bicarbonate in this solution to carbonic acid with a molarity of 0.02M utilizing 1 M HCl.

Moles of sodium bicarbonate = 0.005 mol

Moles of carbonic acd required if solution was 100 mL = 0.002 mol

Moles of sodium bicarbonate reacting with HCl = 0.002 mol

0.002 mol = Molarity of HCl * Volume of HCl

0.002 = 1 M * V

V = 0.002 L = 2 mL

So,

2 mL of HCl should be added.

Molarity of sodium bicarbonate = (0.005 - 0.002) * 1000 / 100 = 0.03 M

Molarity of carbonic acid = 0.02 * 1000 / 100 = 0.02 M

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