Glycine amide hydrochloride (BH+) FM 110.54, pKa 8.20
Glycine amide (B) FM 74.08
b.) How many grams of glycine amide should be added to 1g of glycine amide hydrochloride to give 100ml of solution with pH 8?
I tried using the henderson-hasselbalch equation to give:
8 = 8.2 + log((B/BH+)) though now I'm not sure exactly what I'm trying to solve for.
C.) What would be the pH if the solution in (a) is mixed with 5mL of 0.100M HCl?
Part (a) refers to a solution prepared by dissolving 1g glycine amide hydrochloride plus 1g of glycine amide in 0.100L.
Answer is 8.33
D.) What would be the pH if the solution in (c) is mixed with 10mL of .100M NaOH?
Answer is 8.41
As always, I just need help on finding out how to get to the answers.
Thank you.
To solve these questions, we need to consider the Henderson-Hasselbalch equation, which relates the pH of a solution to the pKa and the ratio of the concentration of the conjugate base (B) to the concentration of the acid (BH+). Let's go step by step through the problem.
b.) How many grams of glycine amide should be added to 1g of glycine amide hydrochloride to give 100ml of solution with pH 8?
Step 1: Using the Henderson-Hasselbalch equation:
pH = pKa + log([B]/[BH+])
Given:
pH = 8.0
pKa = 8.20
[BH+] (concentration of glycine amide hydrochloride) = 1g/110.54g/mol = 0.00904 mol/L
Let's assume the concentration of glycine amide (B) is x mol/L.
So, we can rewrite the Henderson-Hasselbalch equation as:
8 = 8.20 + log(x/0.00904)
Step 2: Solve for x (concentration of glycine amide):
log(x/0.00904) = 8 - 8.20
log(x/0.00904) = -0.20
Step 3: Find the antilog of both sides to solve for x:
x/0.00904 = 10^(-0.20)
x = 0.00904 * 10^(-0.20)
Step 4: Calculate the mass of glycine amide needed to prepare 100 mL of solution:
Mass = concentration (mol/L) * molar mass (g/mol) * volume (L)
Mass = x * 74.08 g/mol * 0.1 L
Substituting the value of x from Step 3:
Mass = (0.00904 * 10^(-0.20)) * 74.08 g/mol * 0.1 L
Now you can calculate Mass.
c.) What would be the pH if the solution in (a) is mixed with 5 mL of 0.100 M HCl?
Step 1: Determine the final volume of the solution after adding 5 mL of HCl.
Final volume = initial volume (100 mL) + volume of HCl (5 mL) = 105 mL
Step 2: Calculate the concentration of added HCl:
Concentration of HCl = 0.100 mol/L (given)
Step 3: Calculate the total moles of glycine amide and HCl in the solution:
Moles of glycine amide = initial concentration (mol/L) * initial volume (L)
Moles of HCl = added concentration (mol/L) * added volume (L)
Step 4: Calculate the new concentration of glycine amide and glycine amide hydrochloride:
New concentration of glycine amide = moles of glycine amide / final volume (L)
New concentration of glycine amide hydrochloride = moles of glycine amide hydrochloride / final volume (L)
Step 5: Use the Henderson-Hasselbalch equation as pH = pKa + log([B]/[BH+]) to find the new pH. Substitute the new concentrations into the equation.
d.) What would be the pH if the solution in (c) is mixed with 10 mL of 0.100 M NaOH?
Follow the same steps as in part (c), but this time, calculate the new concentrations and use the Henderson-Hasselbalch equation to find the new pH.
To solve these problems, you can use the Henderson-Hasselbalch equation, which relates the pH of a solution to the pKa of a weak acid or base and the ratio of its conjugate acid/base concentrations.
b.) To determine how many grams of glycine amide (B) should be added to 1g of glycine amide hydrochloride (BH+) to give a 100ml solution with pH 8, you need to find the ratio of B to BH+ that would result in the desired pH.
First, let's calculate the ratio of B to BH+ using the Henderson-Hasselbalch equation:
pH = pKa + log10(B / BH+)
8 = 8.20 + log10(B / 1)
log10(B) = 8 - 8.20
log10(B) = -0.20
Now, convert the logarithm form back to exponential form, remembering that the antilog of -0.20 is the same as 10 raised to the power of -0.20:
B = 10^(-0.20)
B = 0.63
To find the mass of glycine amide (B) needed to give this concentration ratio in a 100ml solution, use the formula:
Mass of B = (Desired ratio of B/BH+) × Mass of BH+
Mass of B = 0.63 × 1g
Mass of B = 0.63g
So, you would need to add 0.63 grams of glycine amide to 1g of glycine amide hydrochloride to obtain a 100ml solution with pH 8.
c.) To find the pH of the solution prepared by dissolving 1g of glycine amide hydrochloride and 1g of glycine amide in 100ml of solution, and then adding 5ml of 0.100M HCl, you can calculate the new concentration of BH+ and use the Henderson-Hasselbalch equation.
First, calculate the new concentration of BH+:
Initial moles of BH+ = Initial mass of BH+ / Formula weight of BH+
Initial moles of BH+ = 1g / 110.54 g/mol (formula weight of BH+)
Initial moles of BH+ = 0.00904 moles
Moles of BH+ after adding HCl = Initial moles of BH+ - Moles of HCl added
Moles of BH+ after adding HCl = 0.00904 moles - (0.005L × 0.100M) (volume × concentration)
Moles of BH+ after adding HCl = 0.00854 moles
Now, convert moles back to concentration:
[ BH+ ] = Moles of BH+ after adding HCl / Total volume of solution
[ BH+ ] = 0.00854 moles / (100ml + 5ml)
[ BH+ ] = 0.00854 moles / 0.105L (since 100ml + 5ml = 105ml)
pH = pKa + log10([ B ] / [ BH+ ])
8 = 8.2 + log10([ B ] / 0.00854)
Now you need to solve for the concentration of B to calculate the pH. To do this, you'll need to use the mass of B, which is given as 1g in the question. First, calculate the moles of B:
Moles of B = Mass of B / Formula weight of B
Moles of B = 1g / 74.08 g/mol (formula weight of B)
Moles of B = 0.0135 moles
Next, calculate the concentration of B:
[ B ] = Moles of B / Total volume of solution
[ B ] = 0.0135 moles / 0.105L
Substitute the calculated concentrations into the Henderson-Hasselbalch equation:
8 = 8.2 + log10(0.0135 / 0.00854)
Now, solve this equation for the log term and then calculate the logarithm to find the pH.
d.) To find the pH of the solution from part (c) after adding 10ml of 0.100M NaOH, you need to calculate the new concentration of B and use the Henderson-Hasselbalch equation.
First, calculate the new concentration of B:
Moles of NaOH added = Volume of NaOH added × Concentration of NaOH
Moles of NaOH added = 0.010L × 0.100M
Moles of NaOH added = 0.001 moles
Moles of B after adding NaOH = Initial moles of B - Moles of NaOH added
Moles of B after adding NaOH = 0.0135 moles - 0.001 moles
Moles of B after adding NaOH = 0.0125 moles
Now, convert moles back to concentration:
[ B ] = Moles of B after adding NaOH / Total volume of solution
[ B ] = 0.0125 moles / (100ml + 5ml + 10ml) (since 100ml + 5ml + 10ml = 115ml)
pH = pKa + log10([ B ] / [ BH+ ])
pH = 8.2 + log10(0.0125 / 0.00904)
Solve this equation for the log term and then calculate the logarithm to find the pH.
The set up from part A looks like this.
1g/74.08 = moles B = 0.0135
1/110.54 = moles BH^+ = 0.009046
YOu had 100 mL soln for part A; therefore, we have 13.5 mmoles B and 9.046 mmols BH^+
From part c we add 5 mL x 0.1M = 0.5 mmoles.
............B + H^+ ==> BH^+
initial...13.5..0.......9.046
add............0.5.............
change....-0.5..-0.5.....+0.5
equil.....13.0...0.......9.546
pH = 8.2 + log(13.0/9.546) = 8.33
The others are done the same way.