Use this chart to help answer the questions Name 220 nen Procedure: Lab Partner:

Question

Use this chart to help answer the questions

Name 220 nen Procedure: Lab Partner: voolt Using Solution Preparation mix clean glassware, the listed amount of each of the following and making sure the solutions remember to label your make all the solutions components, pH solutions. listed To avoid any confusio ow. Measurement Using a pH meter, measure the pH of each solution record it in the table below. and Solution Composition 10 mL, 0.10M CH3C02H 5mL 0.10M CH3CO2H 5mL, H20 1mL 0.10M CH3CO2H 99mL, H20 5mL 0.10M CH3CO2H 5mL 0.10M HCl 10 mL 0.10M H3PO4 10 mL 0.10M NH3 10 mL 0.10M NH4NO3 0mL 0.10M NH3 50mL 0.10M NH4NO3 10mL solution 8 6mLH20 10mL solution 8 5ml, H20 1mL 0.10M HCI 10mL solution 8 6mL 0,10M HCl 10ml, solution 8 5mL H20 1mL 0.10M NaoH 10 mL, 0.10M CH3CO2H 5mL 0.10M NaoH 13 10mL 0.10M NH4NO3 5ml. 0.10M NaoH 14

Explanation / Answer

I am showing calculations for solution 10 only, rest are similar.

In solution 10, 10 mL of solution 8 ( which is made of 0.05 M each of NH3 and NH4NO3 ) is used, and 1 mL of 0.1 M HCl is also used.

Moles of HCl added = 0.1*0.001 = 10-4 moles

Solution 8 is a buffer of a weak base ( ammonia ) and its salt with a strong acid (HNO3).

pKb of ammonia = 4.65

When HCl is added, more of the base reacts with it to form salt, so effectively the moles of base decreases by the amount of moles of acid added (10-4) and moles of salt increase by this same amount.

In the Henderson Hasselbach equation:

pKb = pOH + log(moles of salt/moles of base)

Initially,

Moles of salt = 0.05*0.008 = 0.0004

Moles of acid = 0.05*0.008 = 0.0004

After addition of HCl,

Moles of salt = 0.0004 + 0.00001 = 0.00041

Moles of acid = 0.0004 - 0.00001 = 0.00039

Putting into eqn we get:

pKb = 4.65 + log(0.00041/0.00039) = 4.67

So, pH = 14 - pOH = 9.33

Solution 11 is solved in a similar way.

In solution 12, addition of NaOH will have a reverse effect. It will increase the moles of acid and decrease the moles of salt by the same amount.

(c)

Diluting the buffer has no impact on its pH, since in the Henderson Hasselbach equation we see that the ratio of moles in the log term will still remain the same even if the buffer is diluted.

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