How much 5M KOH must be added to 1.0 L of 0.1M glycine (pKa 9.6) at pH 9.0 to bring its pH to exactly 10.0?
Use the Henderson Hasselbalch equation twice.
First set of conditions:
pH=9.0
pka=9.6
Solve for the ratio
pH=pka+log[A-/HA]
9.0=9.6+log[A-/HA]
10^(9.0-9.6)=[A-/HA]
0.25=[A-/HA]
Meaning 25% of the solution is protanated, or 0.025 moles is A- and 0.075 moles HA
Second set of conditions
pH=10.0
pka=9.6
Solve for the ratio
pH=pka+log[A-/HA]
10=9.6+log[A-/HA]
10^(10.0-9.6)=[A-/HA]
25.0=[A-/HA]
Meaning 25% of the solution is deprotanated, or 0.025 moles is HA and 0.075 moles A-
I need to go from 0.025 moles of A- to 0.075 moles of A-, and I need to go from 0.075 moles of HA to 0.025 moles of HA. This means that I need 0.05 moles of KOH.
5M KOH=0.05 moles of KOH/L
Solve for volume,
0.05 moles of KOH/5M=L of KOH
Typos---
Change 0.025 to 0.25
change 0.075 to 0.75
change 0.05 to 0.5
Everything is correct.
Well, it seems like you're running a pH balancing circus here! Let's see if we can find the right act to get those pH levels sorted out.
To go from a pH of 9.0 to 10.0, we'll need to increase the pH by 1 unit. Since glycine has a pKa of 9.6, it acts as a buffer in this pH range. In order to increase the pH to 10.0, we'll need to add a strong base to push it up.
To figure out how much 5M KOH we need, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Since we want to increase the pH from 9.0 to 10.0, we can rearrange the equation to solve for the ratio [A-]/[HA]. Plugging in the given values:
10.0 = 9.6 + log([A-]/[HA])
After some mathematical circus tricks, we find that [A-]/[HA] = 10^(10.0-9.6).
Now, let's calculate the concentration of glycine ([HA]) at pH 9.0 using the Henderson-Hasselbalch equation again:
9.0 = 9.6 + log([A-]/[HA])
Solving for the ratio [A-]/[HA], we find that [A-]/[HA] = 10^(9.0-9.6).
With these ratios in hand, we can determine how much 5M KOH is needed to increase the pH to 10.0. Since the volume of the glycine solution is 1.0 L, we need to find the moles of glycine (HA) present and calculate the moles of KOH (A-) necessary to increase the pH.
Once we have the moles of KOH, we can convert it to grams using the molar mass of KOH to get the final answer.
I hope all these calculations don't make your head spin like a juggling clown at a circus!
To answer this question, we need to understand the concept of buffer solutions and how they can be manipulated to achieve a desired pH. Let's break it down step by step.
1. Determine the initial pH of the glycine solution:
The glycine solution has a pH of 9.0, which is below its pKa of 9.6. This means that glycine will predominantly exist in its deprotonated form (conjugate base) in the solution.
2. Determine the desired pH of the solution:
We want to increase the pH of the glycine solution to exactly 10.0. Since the solution already contains a significant amount of glycine in its deprotonated form, we need to add a base to the solution to increase the pH.
3. Identify the base that will react with glycine to increase the pH:
In this case, we can use potassium hydroxide (KOH) as the base to neutralize the glycine and increase the pH. The reaction will be:
KOH + glycine ⇌ glycine^- + H2O
4. Calculate the moles of glycine in the solution:
Given that the initial volume of the solution is 1.0 L and the initial concentration of glycine is 0.1 M, we can calculate the moles of glycine using the formula:
Moles of glycine = initial concentration of glycine × volume of solution
= 0.1 M × 1.0 L
= 0.1 moles
5. Calculate the moles of glycine required to react with KOH and achieve the desired pH:
Since glycine has a pKa of 9.6, it will be half protonated and half deprotonated at pH 9.6. To increase the pH from 9.0 to 10.0, we need the glycine to be almost completely in its deprotonated form. Using the Henderson-Hasselbalch equation, we can calculate the ratio of the concentration of glycine in its deprotonated form ([A^-]) to the concentration of glycine in its protonated form ([HA]):
pH = pKa + log([A^-]/[HA])
Rearranging the equation, we get:
[A^-]/[HA] = 10^(pH - pKa)
= 10^(10 - 9.6)
= 2.51
This means that for every 2.51 moles of glycine deprotonated, there will be 1 mole of glycine protonated. Since we initially have 0.1 moles of glycine, the concentration of deprotonated glycine required is:
[A^-] = (2.51 / 3.51) × 0.1 moles
≈ 0.072 moles
6. Calculate the moles of KOH required to react with glycine:
The molar ratio between KOH and glycine in the reaction is 1:1. Therefore, we need the same number of moles of KOH as the moles of deprotonated glycine:
Moles of KOH = moles of deprotonated glycine
= 0.072 moles
7. Convert moles of KOH to grams:
To convert moles of KOH to grams, we need to know the molar mass of KOH:
Molar mass of KOH = 39.1 g/mol (K) + 16.0 g/mol (O) + 1.0 g/mol (H)
≈ 56.1 g/mol
Therefore, the mass of KOH required is:
Mass of KOH = moles of KOH × molar mass of KOH
= 0.072 moles × 56.1 g/mol
≈ 4.0 grams
In summary, approximately 4.0 grams of KOH should be added to 1.0 L of 0.1 M glycine at pH 9.0 to bring the solution to exactly pH 10.0.