How many grams of sodium acetate( molar mass 82.03 g/mol) must me added to 1.00L of a 0.200M acetic acid solution to form a buffer of 4.20?Ka value for aceic acid is 1.8 x 10^-5.
How do I go about finding the grams from the given information?
http://www.chembuddy.com/?left=pH-calculation&right=pH-buffers-henderson-hasselbalch
You figure concentrations needed from the Henderson Hass equation, then convert concentration to grams, knowing volume and molmass.
I still don't get how to solve my problem because the equation still doesn't make much since to me.
pH = pKa + log(base)/(acid)
You know pH.
Convert Ka to pKa for acetic acid.
You know (acid)
Solve for (base) = sodium acetate.
Convert concentration to grams.
To solve this problem step-by-step:
1. Convert the pH value of the buffer solution to pKa using the formula pKa = -log(Ka). In this case, pKa = -log(1.8 x 10^-5).
2. Calculate the concentration of acetic acid using the formula [acid] = 10^(-pH). In this case, [acid] = 10^(-4.20).
3. Use the Henderson-Hasselbalch equation pH = pKa + log([base]/[acid]) to find the concentration of the sodium acetate ([base]).
4. Rearrange the equation to solve for [base] as follows: [base] = 10^(pH - pKa) * [acid]. Substitute the values calculated in steps 1 and 2 to find [base].
5. Convert the concentration of sodium acetate ([base]) to grams using the formula grams = concentration * volume * molar mass. In this case, volume = 1.00 L and molar mass = 82.03 g/mol.
By following these steps, you can calculate the grams of sodium acetate needed to form the desired buffer.
To solve this problem, first, let's find the pKa value for acetic acid by taking the negative logarithm of the Ka value. Since the given Ka value for acetic acid is 1.8 x 10^-5, the pKa value would be -log(1.8 x 10^-5).
Next, we use the Henderson-Hasselbalch equation, which is pH = pKa + log([base]/[acid]), to find the ratio of [base] to [acid]. Since we know the pH (which is 4.20), and we now know the pKa value, we can rearrange the equation as follows: log([base]/[acid]) = pH - pKa.
Now, substitute the known values into the equation: log([base]/[acid]) = 4.20 - (-log(1.8 x 10^-5)).
Simplify the equation: log([base]/[acid]) = 4.20 + 5(log(1.8)) - 5(log(10)).
Next, convert the logarithms to their respective values: log(1.8) = 0.2553 (approx.) and log(10) = 1.
Substitute these values into the equation: log([base]/[acid]) = 4.20 + 5(0.2553) - 5(1).
Simplify further: log([base]/[acid]) = 4.20 + 1.2765 - 5.
Complete the calculation: log([base]/[acid]) = 0.4765.
Now, we can convert this equation back into exponential form to find the ratio of [base] to [acid]. The equation becomes: 10^(0.4765) = [base]/[acid].
Solve for [base]/[acid]: [base]/[acid] = 3.014.
Since the concentration of acetic acid is given in the problem as 0.200 M, we know that [acid] = 0.200.
Now, we can solve for [base]: [base] = 3.014 * [acid].
Substitute the value of [acid]: [base] = 3.014 * 0.200.
Solve for [base]: [base] = 0.603 M.
We know the volume of the solution is 1.00 L.
Finally, we can convert the concentration to grams using the formula: grams = concentration (M) * volume (L) * molar mass (g/mol).
Substitute the values: grams = 0.603 * 1.00 * 82.03.
Calculate the grams: grams = 49.48 g.
Therefore, you would need to add approximately 49.48 grams of sodium acetate to 1.00 L of the 0.200 M acetic acid solution to form a buffer with a pH of 4.20.
So,
pH = 4.20
pKa = -log(1.8 x 10^-5) = 4.74
(acid) = 0.200M
(base) = (0.200M) / (10^(4.20 - 4.74)) = 0.0045M
Grams of sodium acetate = (0.0045M) * (1.00L) * (82.03 g/mol) = 3.69 g