What volume (in milliliters) of 0.180 M NaOH should be added to a 0.135 L solution of 0.0230 M glycine hydrochloride (pKa1 = 2.350, pKa2 = 9.778) to adjust the pH to 2.73?
Plz help, I'm not sure how to answer the question!
Thank you!!
To answer this question, you'll need to use the Henderson-Hasselbalch equation, which relates the pH of a solution to the pKa and the ratio of the concentrations of weak acid and its conjugate base.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
Where:
- pH is the desired pH (2.73 in this case)
- pKa is the acid dissociation constant of the weak acid (glycine hydrochloride), pKa1 in this case (2.350)
- [A-] is the concentration of the conjugate base (glycine)
- [HA] is the concentration of the weak acid (glycine hydrochloride)
First, let's determine the concentrations of glycine and glycine hydrochloride in the 0.135 L solution.
To do that, we use the formula for molarity:
Molarity (M) = moles of solute ÷ volume of solution (in liters)
For glycine hydrochloride:
Molarity = 0.0230 M
So, the moles of glycine hydrochloride in the solution are:
moles of glycine hydrochloride = Molarity × volume (in liters)
= 0.0230 M × 0.135 L
≈ 0.00310 moles of glycine hydrochloride
Now, let's calculate the moles of glycine in the solution based on the given pKa1 value.
We use the Henderson-Hasselbalch equation to find the ratio of [A-]/[HA]:
2.73 = 2.350 + log([A-]/[HA])
Rearranging the equation, we get:
log([A-]/[HA]) = 2.73 - 2.350
log([A-]/[HA]) = 0.380
Now, we can use the antilog function to find the ratio of [A-]/[HA]:
[A-]/[HA] = 10^(0.380)
[A-]/[HA] ≈ 2.380
Since the ratio [A-]/[HA] is approximately equal to 2.380, we can assume that the concentration of glycine is approximately two times the concentration of glycine hydrochloride in the solution.
So, the moles of glycine in the solution are:
moles of glycine = 2 × 0.00310 moles of glycine hydrochloride
≈ 0.00620 moles of glycine
Now that we have the moles of glycine and glycine hydrochloride, we can calculate the volume of 0.180 M NaOH needed to adjust the pH.
Since glycine is a weak acid, it reacts with NaOH in a 1:1 ratio. This means that for every mole of glycine, we need one mole of NaOH.
Therefore, moles of NaOH needed = moles of glycine
Now, let's determine the volume of 0.180 M NaOH needed:
Volume of NaOH (in liters) = moles of NaOH ÷ Molarity
= moles of glycine ÷ 0.180 M
≈ 0.00620 moles ÷ 0.180 M
≈ 0.0344 L
Finally, convert the volume to milliliters:
Volume of NaOH (in milliliters) = 0.0344 L × 1000 mL/L
≈ 34.4 mL
Therefore, approximately 34.4 milliliters of 0.180 M NaOH should be added to the solution to adjust the pH to 2.73.