What volume (in milliliters) of 0.170 M NaOH should be added to a 0.140 L solution of 0.021 M glycine hydrochloride (pKa1 = 2.350, pKa2 = 9.778) to adjust the pH to 2.84?
I tried using the Henderson- Hasselbalch equation but I am confused about what to do with the answer to apply it to the question.
After solving 2.84=2.350+log(A-/HA), I got
A- = 0.0158659
HA= 0.0051341
To solve this problem, you have correctly used the Henderson-Hasselbalch equation:
pH = pKa + log(A-/HA)
You have found the values for A- and HA. A- represents the conjugate base of the acid (glycine hydrochloride in this case) and HA represents the acid itself.
Now, we need to determine the amount of NaOH needed to adjust the pH to 2.84. Since NaOH is a strong base, it will fully dissociate in water to Na+ and OH-. The OH- ions will react with the HA (glycine hydrochloride) to form water and generate the conjugate base (A-).
The balanced chemical equation for this reaction is:
HA + OH- -> A- + H2O
The stoichiometry of the reaction tells us that one mole of HA reacts with one mole of OH-, so the amount of NaOH needed can be calculated based on the moles of HA present in the solution.
To calculate the initial moles of HA present in the solution, we use the molarity and volume:
moles of HA = Molarity of HA × Volume of HA solution
moles of HA = 0.021 M × 0.140 L = 0.00294 mol
Since the reaction between HA and OH- is 1:1, we need 0.00294 moles of OH- to react completely with HA and generate A-.
Now, we need to calculate the volume of 0.170 M NaOH solution needed to provide 0.00294 moles of OH-.
moles of NaOH = Molarity of NaOH × Volume of NaOH solution
0.00294 mol = 0.170 M × Volume of NaOH solution
Solving for the Volume of NaOH solution:
Volume of NaOH solution = 0.00294 mol / 0.170 M
Volume of NaOH solution = 0.0173 L = 17.3 mL
Therefore, you would need to add 17.3 mL of 0.170 M NaOH to adjust the pH of the solution to 2.84.