Posted by **Cassandra** on Monday, April 2, 2012 at 11:39pm.

Tris is an abbreviation for tris (hydroxymethyl) aminomethane, which is a weak base. Calculate how many grams of tris (mw=121.14g/mol) are needed to make 25mL of 0.10M tris solution. Calculate how many mL of 0.25M HCl are needed to convert half of the moles of tris present in 25mL of 0.10 M tris solution to protonate tris (trisH^+). The relevant hydrolysis equilibrium is tris+ H^+ -> Htris^+

- Chemsitry -
**DrBob222**, Tuesday, April 3, 2012 at 12:03am
You want 25 mL of 0.1M; that is mols = g/molar mass = ?

You want to convert half of that to the acid; therefore, 1/2 x mols tris = ?

Now, M HCl = mols HCl/L HCl.

You kow M and mols, solve for L HCl and convert to mL.

- Chemsitry -
**Cassandra**, Tuesday, April 3, 2012 at 12:08am
A chemist needs to make 50 mL buffer solution with a pH value of 7.25 using the acid/conjugate base pair of KH2PO4 and Na2HPO4 (which has an effective pKa = 6.86). The buffer solution needs to be made so that the total concentration of buffer is 0.10 M (that is, [acid] + [base] = 0.10 M). Calculate the mass of KH PO and Na HPO needed using the following formula:

10^(7.25−𝑝𝐾𝑎) = (0.10 M×0.050 L−mol HA)/mol HA

Solve for mol acid and then use the molecular weight of the acid to calculate the mass (grams) of acid. Then solve for mol base using mol A- + mol HA = 0.10 M × 0.05 L and calculate the mass (grams) of base.

- Chemsitry -
**Cassandra**, Tuesday, April 3, 2012 at 12:19am
Starting out with 50 mL of 0.20 M NaHCO3, calculate how many mL of 0.50 M NaOH solution to add to make 100 mL of approximately 0.10 M (total) buffer solution with a pH of 10.35. By adding NaOH, some of the NaHCO3 gets converted to the conjugate Na2CO3. This conjugate acid/base pair has an effective pKa of 10.00. Calculate mL of 0.50 M NaOH needed from the following calculations:

10.35-pK =log[(mol CO3^2-)/(mol HCO3^-)]

There is initially Na2CO3 present in the 50 mL of 0.20 M NaHCO3 solution. By adding some NaOH, you will create some Na2CO2 in the solution and also decrease the mol of NaHCO2. In fact, mol NaOH added = mol Na CO in buffer solution. Therefore,

10^(10.35−𝑝𝐾𝑎)= [(mol NaOH added)/(0.20 M×0.050 L−mol NaOH added)]

Solve for mol NaOH added in the above equation. Then, use the concentration of the NaOH (0.50 mol/L) to calculate the calculate the mL of NaOH necessary to prepare the buffersolution.

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