Calculate the pH of 1.0 Lof original buffer solution which contain 0.50 M in NH3

Question

Calculate the pH of 1.0 Lof original buffer solution which contain 0.50 M in NH3 and 0.20 M in NH4Cl ( for ammonia, pKb=4.75) upon addition of, a) 0.010 mol of solid NaOH
b) 30.0 mL of 1.0 M HCl
Pls show all chemical equations.
Thanks Calculate the pH of 1.0 Lof original buffer solution which contain 0.50 M in NH3 and 0.20 M in NH4Cl ( for ammonia, pKb=4.75) upon addition of, a) 0.010 mol of solid NaOH
b) 30.0 mL of 1.0 M HCl
Pls show all chemical equations.
Thanks a) 0.010 mol of solid NaOH
b) 30.0 mL of 1.0 M HCl
Pls show all chemical equations.
Thanks

Explanation / Answer

a) 0.010 mole of solid NaOH is added

Given-Volume of original buffer solution = 1L

0.50 M in NH3

0.20 M in NH4Cl

For ammonia   pKb = 4.75

Step 1- Calculate initial moles of NH3 and NH4Cl

NH3 moles = 0.5 * 1.0 = 0.50 moles

NH4Cl moles = 0.20 * 1.00 = 0.20 moles

Step 2-Re-calculate the molar amounts

Chemical Buffer tends to slow-down the pH-influences exercited by Acid/Bases.

NH4Cl (aq) + NaOH (aq) <---> NH3 (aq) + NaCl (aq) + H2O (aq)

Hence NH3 is forming while NH4Cl 1s sinking

So, I re-calculate the molar amounts

NH3 moles = 0.50 + 0.010 = 0.51 mole

NH4Cl moles = 0.20 - 0.010 = 0.19 mole

Step- 3 Calculation of PH

According to Henderson-Hassel Bach equation -

pH = 14 - pKb + log[NH3 moles]/[NH4Cl moles])

pH = 14.00 - 4.75 + log([0.51]/[0.19]) = 9.67

pH = 14.00 - 4.75 + 0.43 = 9.68

b) 30.0 mL of 1.0 M HCl is added

Step-1 Calculate the moles of HCl

Moles of HCl = 1.0 M x 0.030L = 0.03 moles

Step-2 calculate the moles of NH3

When we add a small amount of Hydrochloric Acid, it reacts with Ammonia

NH3 (aq) + HCl(aq) <---> NH4Cl(aq)

Hence NH3 is sinking while NH4Cl is forming.

So, I re-calculate the molar amounts

0.03 moles of HCl will react with 0.03 moles of NH3

NH3 moles =0.50 - 0.03 = 0.47 mole

NH4Cl moles = 0.20 + 0.03 = 0.23 mole

Step- 3 Calculation of PH

According to Henderson-Hassel Bach equation -

pH = 14.00 - 4.75 + log([NH3]/[NH4Cl])

pH = 14.00 - 4.75 + log 0.47/0.23

pH = 14.00 - 4.75 + 0.31 = 9.56

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