calculate the pH after 0.02 mol NaOH is added to 1.00 L of .1 M sodium propanoate NaC3H5O2
I THINK all you have to do here is to calculate the pH of the 0.02 M NaOH but just for good measure I would soolve for the contribution from the Na salt. If it is large enough add it to that from hydrolysis of the Na salt.
To calculate the pH after adding NaOH to a solution of sodium propanoate, we need to consider the reaction that occurs between NaOH and sodium propanoate.
The reaction equation is:
NaOH + C3H5O2Na → C3H5O2H + NaOH
In this reaction, NaOH reacts with the sodium propanoate (C3H5O2Na) to form propanoic acid (C3H5O2H) and more NaOH.
Now, let's calculate the moles of sodium propanoate in the 1.00 L solution:
moles of C3H5O2Na = molarity x volume
= 0.1 M x 1.00 L
= 0.1 mol
Since the ratio between NaOH and C3H5O2Na in the reaction is 1:1, adding 0.02 mol of NaOH will result in the complete neutralization of 0.02 mol of C3H5O2Na.
Therefore, the moles of C3H5O2Na remaining in the solution will be:
moles of C3H5O2Na remaining = initial moles - moles of NaOH added
= 0.1 mol - 0.02 mol
= 0.08 mol
After the neutralization reaction, the remaining sodium propanoate (C3H5O2Na) will be in equilibrium with its conjugate acid form, propanoic acid (C3H5O2H).
To calculate the concentration of propanoic acid, we divide the moles of propanoic acid by the volume of the solution:
concentration of C3H5O2H = moles of C3H5O2H / volume
= 0.08 mol / 1.00 L
= 0.08 M
Now, we can determine the pKa value of propanoic acid:
pKa = -log(Ka)
The Ka value for propanoic acid is 1.34 x 10^-5 (from a chemistry reference).
Now, let's calculate the pKa value:
pKa = -log(1.34 x 10^-5)
= 4.87
Since propanoic acid is a weak acid, we can use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log([A-] / [HA])
Where [A-] is the concentration of the conjugate base (sodium propanoate) and [HA] is the concentration of the weak acid (propanoic acid).
Plugging in the values:
pH = 4.87 + log(0.08 / 0.08)
= 4.87 + log(1)
= 4.87 + 0
= 4.87
Therefore, the pH after adding 0.02 mol NaOH to 1.00 L of 0.1 M sodium propanoate NaC3H5O2 is approximately 4.87.