Question

7. What mass of solid NH4Cl and what volume of 1.00 mol-L¹ NaOH solution should be used to

prepare 1 L of a buffer solution of pH 9.00? Suppose the overall concentration of the buffer is 0.125
mol-L¹. (Answer V = 45 mL)

Answer 1

The pKa of NH4Cl is 9.25. Therefore, the pH of the buffer can be calculated using the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

where [A-]/[HA] is the ratio of the concentration of the conjugate base to the concentration of the weak acid. Since NH4Cl is a salt of a weak acid (NH4+) and a strong base (Cl-), the weak acid in this case is NH4+.

Rearranging the Henderson-Hasselbalch equation gives:

[A-]/[HA] = 10^(pH - pKa)

Substituting the given values into the equation:

[A-]/[HA] = 10^(9.00 - 9.25) = 0.562

Since the overall concentration of the buffer is 0.125 mol-L¹, we can set up the following two equations:

[A-] + [HA] = 0.125 mol-L¹

[V1] [C1] = [V2] [C2]

where V1 is the volume of NaOH solution, C1 is the concentration of NaOH solution, V2 is the total volume of the buffer solution (1 L), and C2 is the concentration of NH4Cl.

Since NH4Cl is a 1:1 electrolyte, [A-] = [NH3] and [HA] = [NH4+]. Therefore, we can rewrite the first equation as:

[NH3] + [NH4+] = 0.125 mol-L¹

Substituting [A-]/[HA] = 0.562 into the equation gives:

[NH3] = 0.562 [NH4+]

Substituting this into the equation [NH3] + [NH4+] = 0.125 mol-L¹ gives:

[NH4+] = 0.125 / (1 + 0.562) = 0.0517 mol-L¹

[C2] = 0.0517 mol-L¹

The molar mass of NH4Cl is 53.49 g-mol¹. Therefore, the mass of NH4Cl needed to prepare 1 L of a 0.0517 mol-L¹ solution is:

m = C × V × M

where m is the mass of NH4Cl, C is the concentration of NH4Cl, V is the volume of

Answer 2

STEP BY STEP SOLUTION :

To prepare a buffer solution of pH 9.00, we need to use the Henderson-Hasselbalch equation:

• pH = pKa + log([A-]/[HA])

Where [A-]/[HA] is the ratio of the concentrations of the conjugate base and acid of the buffer, respectively. Since we are given the pH and the overall concentration of the buffer, we can solve for the ratio [A-]/[HA]:

• 9.00 = pKa + log([A-]/[HA])
• pKa = 9.25 (the pKa of NH4Cl)
• 9.00 = 9.25 + log([A-]/[HA])
• log([A-]/[HA]) = -0.25
• [A-]/[HA] = 0.56

Next, we can use the definition of the concentration of a solution to find the concentration of NH4Cl needed to make a 0.125 mol-L^-1 buffer solution:

0.125 mol-L^-1 = [NH4Cl] + [NaOH]

Since the NaOH solution is 1.00 mol-L^-1, we can assume that the contribution of NaOH to the total concentration of the buffer is negligible, and so:

• 0.125 mol-L^-1 = [NH4Cl]

Finally, we can use the molar mass of NH4Cl to find the mass of NH4Cl needed to prepare 1 L of the buffer solution:

• mass NH4Cl = molar mass * moles
• mass NH4Cl = (14.01 g-mol^-1 + 1.01 g-mol^-1 + 35.45 g-mol^-1) * 0.125 mol
• mass NH4Cl = 6.63 g

So we need to use 6.63 g of NH4Cl and enough volume of 1.00 mol-L^-1 NaOH solution to make a total volume of 1 L. To find the volume of NaOH solution needed, we can use the definition of molarity:

• molality = moles of solute / volume of solution (in liters)

Rearranging this equation, we get:

• volume of solution = moles of solute / molality

Since we are adding NaOH solution to NH4Cl to make a total volume of 1 L, the molality of NaOH solution is also 0.125 mol-L^-1. Therefore:

• volume of NaOH solution = moles of NaOH / molality of NaOH
• volume of NaOH solution = (1 L - volume of NH4Cl solution) * 0.125 mol-L^-1

Substituting the values we know:

• volume of NaOH solution = (1 L - 0.45 L) * 0.125 mol-L^-1
• volume of NaOH solution = 0.056 L = 56 mL

So we need to use 6.63 g of NH4Cl and 56 mL of 1.00 mol-L^-1 NaOH solution to prepare 1 L of a buffer solution of pH 9.00.