So I already posted this question earlier today and got a reply but now that Im working through it, its not working.
What mass of solid sodium formate (of MW
68.01) must be added to 130 mL of 0.63 mol/L
formic acid (HCOOH) to make a buffer solution
having a pH of 3.68? Ka = 0.00018 for
HCOOH.
Answer in units of g.
Using pH = pKa + log [salt]/[acid] formula.
To find pKa, I did -log(.00018) = 3.745
so...
3.68 = 3.745 + log [salt]/[.63]
-.0647 = log [salt]/[.63]
.86153 = [salt]/[.63]
[salt] = .54276
Therefore mass should be [.54276] x 68.01 = 36.91
But Dr Bob said it should be around 5 so im not sure what ive done wrong
You didn't finish. You're right with 36.91 grams but that is for 1,000 mL of solution (remember M = mols/L). You want grams in 130 mL and 36.91 x 130/1000 is very close to 5g but you can work out the exact number.
By the way, I usually follow up at the original post in case a student has a follow up question. Had you showed your work there I would have caught it faster than looking at a new post. However, I'm glad you worked it out. For another thought, if you had used mols and not mols/L for the (acid) you would not have needed that last step I showed of 36.91 x (130/1000). To be honest, however, most profs prefer that you use M or mols/L. Technically, it is correct, but it makes the problem longer. The answer is the same no matter how you do it.
Good to know, thank you so much!
To determine the mass of solid sodium formate required to make the buffer solution, we need to go through the problem step by step.
Step 1: Calculate the pKa value from the given Ka value.
Given: Ka = 0.00018
To find pKa, use the formula pKa = -log(Ka).
pKa = -log(0.00018) ≈ 3.74
Step 2: Apply the Henderson-Hasselbalch equation.
The Henderson-Hasselbalch equation is:
pH = pKa + log([salt]/[acid])
We are given:
pH = 3.68
pKa = 3.74
[acid] (concentration of formic acid) = 0.63 mol/L
Let's solve for [salt] (concentration of sodium formate):
3.68 = 3.74 + log([salt]/0.63)
-0.06 = log([salt]/0.63)
Step 3: Solve for [salt].
To find [salt], we need to get rid of the logarithm. Convert the equation to an exponential form:
10^(-0.06) = [salt]/0.63
[salt]/0.63 ≈ 0.86153 (using a calculator)
[salt] ≈ 0.86153 * 0.63 ≈ 0.543 g/mL
Step 4: Calculate the mass of sodium formate.
The molar mass of sodium formate (HCOONa) is given as 68.01 g/mol.
Mass of sodium formate = [salt concentration] * molar mass
Mass of sodium formate = 0.543 * 68.01 ≈ 37 g
Therefore, the correct mass of solid sodium formate required to make the buffer solution is approximately 37 g, which is close to the value obtained from your calculation. It seems that you made a slight error during the calculation. Make sure to double-check your calculations and units to avoid any mistakes.