What is the pH of buffer system prepared by dissolving 10.70 grams of NH4Cl and

Question

What is the pH of buffer system prepared by dissolving 10.70 grams of NH4Cl and 25.00 mL of 12 M NH3 in enough water to make 1.000 L of solution? Kb = 1.80 x 10^-5 for NH3 Calculate the change in pH after addition of 50.0 mL of 0.15 M NaOH. What is the pH of buffer system prepared by dissolving 10.70 grams of NH4Cl and 25.00 mL of 12 M NH3 in enough water to make 1.000 L of solution? Kb = 1.80 x 10^-5 for NH3 Calculate the change in pH after addition of 50.0 mL of 0.15 M NaOH. Calculate the change in pH after addition of 50.0 mL of 0.15 M NaOH.

Explanation / Answer

We are given the volume and molarity of ammonia, mass of ammonium chloride. We can see that ammonium chloride and ammonia form buffer. And to calculate the pH of the buffer we can use the Henderson hasselbalch equation.

Calculation of concentration NH4Cl

[NH4Cl] = mol / volume in L

            = (mass in g / molar mass ) / volume in L

            = ( 10.70 g / 53.491 g per mol) / 1.0 L = 0.20 M

[NH3] = ( Volume in L x molarity ) / 1.0 L

            = ( 0.025 L x 12 M ) / 1.0

            =0.30 M

Henderson equation

pOH = - log ( kb ) + log ( [NH4Cl]/[NH3])

Lets plug given and calculated values,

pOH = - Log ( 1.8 E-5) + log ( 0.2/ 0.3)

= 4.57

pH = 14 – pOH = 14 – 4.57 = 9.43

so the pH of the buffer would be = 9.43

Calculation of pH change when we add NaOH

Calculation of moles of NaOH

n NaOH = 0.050 L x 0.15 M = 0.0075 mol

We know NaOH is going to react with ammonium chloride. Lets show the reaction.

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

From this reaction we can say that, resultant moles of ammonium chloride will be the original moles of it – mole s of sodium hydroxide.

And resultant moles of ammonia would be original moles of ammonia + moles of NaOH

Resultant moles calculation

Mol NH4Cl = 0.200 – 0.0075 = .192534 mol NH4Cl = [NH4Cl]

Mol NH3 = 0.300 + 0.0075 = 0.3075 mol NH3 = [NH3]

We can use the Henderson equation again to calculate the pH of this solution

pOH = - log ( 1.8 E-5 ) + log ( 0.192534 /0.3075 )

pOH = 4.54

pH = 14 - 4.54 = 9.45

Change in pH = 9.45 -9.43 = 0.02

So the delta pH = 0.02

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