Hi, I'm working out this question for homework, but i'm a little stuck.

How many g of sodium hydrogen carbonate (NaHCO3) and 4.836 g of sodium carbonate must be dissolved in 750 mL of H20 in order to make a buffer with a pH of 10.27??

the ka of sodium hydrogen carbonate is 5.6 x 10 ^11

here is what i did:
4.836 g NaCO3(1 mol/105.99 NaCO3) = 0.04563 mols NaCO3

0.04536 mols/.750 L = 0.0608 M Na2CO3

Am i on the right track? What do i do next? Thanks, I'm completely at a loss for what to do next.

Use the Henderson-Hasselbalch equation.

pH = pKa + log[(base)/(acid)]
You have done the Na2CO3 correctly. That goes in for base. Plug in the other numbers and calculate M for the acid (NaHCO3). Knowing M and volume and molar mass you can calculate grams. By the way, you have Ka = 5.6 x 10^11. I think that's 5.6 x 10^-11 but check me out on that.

thank you, i understand now! that was so much clearer than my instructor's explanation.

You're on the right track! Let's break down the problem further and calculate the amount of sodium hydrogen carbonate (NaHCO3) needed next.

First, let's find the number of moles of NaHCO3 required to make the buffer solution.

To do this, we need to calculate the concentration of HCO3- ions in the buffer solution. The pH of a buffer is determined by the ratio of the concentration of the conjugate base (HCO3-) to the weak acid (H2CO3) present in the buffer.

The Henderson-Hasselbalch equation for pH is:

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

Where:
- pH is the desired pH (given as 10.27)
- pKa is the -log(Ka) value of NaHCO3 (given as 11 - log(5.6 x 10^11) = -log(5.6) = 3.25)
- [A-] is the concentration of HCO3- ions (which we want to calculate)
- [HA] is the concentration of H2CO3 (which we can assume to be equal to the concentration of NaHCO3, as the latter fully dissociates in water)

Now, let's rearrange the formula to solve for [A-]:

[A-] = 10^(pH - pKa) * [HA]

[A-] = 10^(10.27 - 3.25) * [HA]

[A-] = 10^7.02 * [HA]

Since the concentration of [A-] is the same as the concentration of NaHCO3 in the solution, we can write:

[NaHCO3] = 10^7.02 * [NaHCO3]

Next, we solve for the concentration of NaHCO3 using the molar mass:

0.04563 mol Na2CO3 * (1 mol NaHCO3 / 2 mol Na2CO3) = 0.02281 mol NaHCO3

Now, we convert the moles of NaHCO3 to grams:

0.02281 mol NaHCO3 * (84.01 g/mol NaHCO3) = 1.917 g NaHCO3

Therefore, you will need approximately 1.917 grams of sodium hydrogen carbonate (NaHCO3) and 4.836 grams of sodium carbonate (Na2CO3) to dissolve in the 750 mL of water in order to make a buffer with a pH of 10.27.