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
I tried converting the pH to H+ and using the Henderson-Hasselbalch equation and got 6.465232859g, which is wrong apparently. A comprehensive explanation would be appreciated!
HH equation is
pH = pKa + log (base)/(acid)
Calculate pKa for listed Ka.
Plug in pKa and pH and mols acid into the HH equation. Solve for (base) = mols Naformate. Convert mols Naformate to grams and you have it. I ran through a rough approx and I think grams Naformate = about 5 g.
Well, it seems like you've already made an attempt, but let's see if we can figure this out together.
To make a buffer solution, we need a weak acid and its conjugate base. In this case, formic acid (HCOOH) is the weak acid and sodium formate (NaHCOO) is its conjugate base.
The Henderson-Hasselbalch equation is quite handy in calculating the pH of a buffer solution:
pH = pKa + log([A-]/[HA])
where pKa is the negative logarithm of the acid dissociation constant (Ka), [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.
First, let's calculate the concentration of formic acid ([HA]) in mol/L. We know the volume (130 mL) and the molarity (0.63 mol/L), so we can use the equation:
[HA] = (0.63 mol/L) × (130 mL) / 1000 mL/L = 0.0819 mol
Now we can substitute this value into the Henderson-Hasselbalch equation:
3.68 = -log(0.00018) + log([A-]/0.0819)
Simplifying:
3.68 + log(0.00018) = log([A-]/0.0819)
Using logarithmic properties:
log([A-]/0.0819) = 3.68 + log(0.00018)
log([A-]/0.0819) = 3.68 + (-3.74)
log([A-]/0.0819) = -0.06
Now, we can solve for [A-]:
[A-]/0.0819 = 10^(-0.06)
[A-] = (10^(-0.06)) × 0.0819
[A-] = 0.0916 mol
Now, keeping in mind that [A-] represents the concentration of sodium formate, we can calculate the mass using its molar mass (68.01 g/mol):
Mass = [A-] × molar mass
Mass = 0.0916 mol × 68.01 g/mol
Mass ≈ 6.23 g
So, the mass of solid sodium formate that must be added to the solution is approximately 6.23 grams.
I hope this helps! And remember, if math ever gets too boring, it's always good to add a pinch of humor to the equation.
To determine the mass of solid sodium formate needed to make a buffer solution with a pH of 3.68, we can follow these steps:
Step 1: Write the balanced chemical equation for the dissociation of formic acid (HCOOH):
HCOOH ⇌ H+ + HCOO-
Step 2: Determine the concentrations of H+ and HCOO- in the buffer solution using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Given:
pH = 3.68
Ka = 0.00018
Since the concentration of the formic acid, [HA], is not provided, we will need to calculate it using the initial concentration and the dissociation equilibrium. The concentration of formic acid can be determined from the given molarity (0.63 mol/L) and volume (130 mL) as follows:
[HA] = (0.63 mol/L) x (0.130 L) = 0.0819 mol
Now, let's calculate the concentration of the formate ion, [A-]:
pH = pKa + log([A-]/[HA])
3.68 = -log(0.00018) + log([A-]/0.0819)
3.68 + log(0.00018) = log([A-]/0.0819)
log([A-]/0.0819) = 3.68 + log(0.00018)
log([A-]/0.0819) = 3.68 + (-3.74)
log([A-]/0.0819) = -0.06
Now, we can find the value of [A-]:
[A-]/0.0819 = 10^(-0.06)
[A-] = 0.0819 x 10^(-0.06)
[A-] ≈ 0.0819 x 0.9405
[A-] ≈ 0.0771 mol
Step 3: Calculate the moles of sodium formate required to react with the formic acid:
Looking at the balanced chemical equation, we can see that for every mole of formic acid that reacts, one mole of sodium formate is required. So, the moles of sodium formate required will be equal to the moles of formic acid:
moles of sodium formate = 0.0771 mol
Step 4: Convert moles of sodium formate to grams:
To convert moles to grams, we need to multiply the number of moles by the molar mass of sodium formate (MW = 68.01 g/mol):
mass of sodium formate = 0.0771 mol x 68.01 g/mol
mass of sodium formate ≈ 5.25 g
Therefore, the mass of solid sodium formate that must be added to the solution is approximately 5.25 grams.
Sure! Let's solve this step by step.
First, let's understand the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
where:
- pH is the measure of acidity or alkalinity of a solution.
- pKa is the negative logarithm of the acid dissociation constant (Ka) of the acid.
- [A-] is the concentration of the conjugate base.
- [HA] is the concentration of the acid.
In this case, the acid is formic acid (HCOOH), and we want to make a buffer solution with a pH of 3.68. The dissociation constant (Ka) for formic acid is given as 0.00018.
Step 1: Convert pH to [H+]
Since pH = -log[H+], we can rearrange the equation to find [H+].
[H+] = 10^(-pH)
[H+] = 10^(-3.68)
[H+] ≈ 2.14 x 10^(-4) mol/L
Step 2: Determine the concentration of formate ion ([A-])
From the Henderson-Hasselbalch equation, we know that [A-]/[HA] = 10^(pH - pKa). Rearranging this equation, we get:
[A-] = [HA] x 10^(pH - pKa)
Step 3: Calculate the concentration of formic acid ([HA])
The concentration of formic acid ([HA]) is given as 0.63 mol/L in 130 mL of solution.
[HA] = (0.63 mol/L) x (0.130 L) = 0.0819 mol
Step 4: Calculate the concentration of formate ion ([A-])
[A-] = (0.0819 mol) x 10^(3.68 - (-0.7449)) [using pKa = -log(Ka)]
[A-] ≈ 0.0819 mol x 10^4.4249
Step 5: Calculate the mass of sodium formate (NaHCOO)
To find the mass of sodium formate required, we need to multiply the number of moles of formate ion ([A-]) by the molar mass of sodium formate (NaHCOO, MW = 68.01 g/mol).
Mass of NaHCOO = [A-] x MW
Mass of NaHCOO ≈ 0.0819 mol x 68.01 g/mol
Mass of NaHCOO ≈ 5.57 g
Therefore, approximately 5.57 grams of solid sodium formate (NaHCOO) must be added to 130 mL of 0.63 mol/L formic acid (HCOOH) to make a buffer solution with a pH of 3.68.
I hope this explanation helps you understand the steps involved in solving this problem!