Pls, help me answer to question number 2! thank you in advance Above, is just a

Question

Pls, help me answer to question number 2! thank you in advance

Above, is just a graph from the data table of the experiment. ?

Report Sheet and Data Analysis DATA TABLE Buffer A Buffer B I, So Mass of NaC2Hs02 used to prepare buffer (g) Volume of buffer prepared (mL) Molar concentration of HC2Hs02 in buffer (M) Initial pH of buffer Volume of 0.5 M NaOH to raise pH by 2 units (m) .3 Volume of 0.5 M HCI to lower pH by 2 units (mL) Volume of 0.5 M NaOH at equivalence point (mL) | | 5 ISg 100.0 100.0 0.1 1.0 4.31 42 ho | I a·5

Explanation / Answer

Buffer capacity can be defined as the amount of acid or base that is added to 1L of a buffer solution inorder to change its pH by 1 unit.

It is expressed in terms of moles/L and given by the following formula:

Buffer capacity = moles of acid or base/change in pH * volume of buffer taken

NOTE: As per the given data, the amount of buffer prepared is 100 ml. I am assuming that you are taking a volume of 25ml.

Buffer A

1) Addition of 2.3 ml of 0.5 M NaOH raises the pH by 2 units

Moles of NaOH added = Molarity * volume in Liters = 0.5 * 0.0023 = 0.000115 moles

Buffer capacity = 0.000115 moles / 2 * 0.025 L = 0.023 moles/L

2) Addition of 0.7 ml of 0.5 M HCl lowers the pH by 2 units

Moles of HCl added = 0.5 * 0.0007 = 0.00035 moles

Buffer capacity = 0.00035 moles /2*0.025 L = 0.007 moles/L

Buffer B

1) Addition of 20.0 ml of 0.5 M NaOH raises the pH by 2 units

Moles of NaOH added = Molarity * volume in Liters = 0.5 * 0.02 = 0.01 moles

Buffer capacity = 0.01 moles / 2 * 0.025 L = 0.2 moles/L

2) Addition of 4.0 ml of 0.5 M HCl lowers the pH by 2 units

Moles of HCl added = 0.5 * 0.004 = 0.002 moles

Buffer capacity = 0.002 moles /2*0.025 L = 0.04 moles/L

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