Respuesta :
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:
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
