Question

6. Calculate the pH of the 0.20 M NH3/0.20 M NH4Cl buffer. What is the pH of the buffer after the addition of 10.0 mL of 0.10 M HCl to 65.0 mL of the buffer

Answer 1

Answer:

Check explanation

Explanation:

The addition of hydrochloric acid, HCl(a strong acid) to the buffer will cause a decease in pH. Although, this changes in PH will not be significant since the role of a buffer is to resist significant changes in pH that result from the addition of strong acid or strong bases.

Step one:

HCl + NH3 --------------> NH4^+ + Cl^- --------------------------------------------(1).

From the equation (1) of reaction above we can reduce that The reaction uses one mole of hydrochloric acid and one mole of ammonia.

10mL × 0.1M HCl= 1 mmols HCl.

65.0mL × 0.20 M= 13 mmols of NH3.

65.0mL × 0.20 M = 13 mmols NH4^+1

Hence, pH= pka + log(13/14).

pH= -6.3 + log 0.93.

pH= -6.3+ (-0.032).

pH= -6.332.

Answer 2

Answer:

The pH before adding HCl is 9.25

The pH after adding HCl is 9.18

Explanation:

Step 1: Data given

Molarity of NH3 = 0.20 M

Molarity of NH4Cl = 0.20

Step 2: Calculate pH of the buffer

pH = pKa + log ([NH3]/[NH4+])

pH = 9.25 + log (0.20/0.20)

pH = 9.25

What is the pH of the buffer after the addition of 10.0 mL of 0.10 M HCl to 65.0 mL of the buffer

Step 1: Calculate moles of HCl added

Moles HCl = 0.01 L * 0.10 M = 0.001 moles

Step 2: Calculate initial moles of NH3

moles of NH3  = 0.2 M * 0.065L = 0.013  moles

 Step 3: The balanced equation

HCl + NH3 → NH4Cl

Step 4: Calculate moles after addition of HCl

moles NH3 after HCl = 0.013 - 0.001 = 0.012 moles NH3

moles NH4 initially present = 0.2mol/L * 0.065L = 0.013 moles

moles NH4 after HCl = 0.013 + 0.001 = 0.014 moles NH4Cl

Step 5: Calculate molarity

Final volume = 10 ml + 65 ml = 75 ml = 0.075 L

Final [NH3] = 0.012 mol/0.075L = 0.16 M  

Final [NH4Cl] = 0.014mol/0.075L = 0.187 M

Step 6: Calculate pH

pH = pKa + log [NH3]/[NH4+] = 9.25 + log 0.16/0.187

pH = 9.18

The pH of the new solution is 9.18