A buffer made up of equal volumes (433.5 mL) of each of 0.671 M H A and 0.635 M A minus. What is p H of the buffer after addition of 6.12 mL of 0.171 M N a O H? K a of H A is 1.480E-5
5.
A buffer made up of equal volumes (433.5 mL) of each of 0.671 M H A and 0.635 M A minus. What is p H of the buffer after addition of 6.12 mL of 0.171 M N a O H? K a of H A is 1.480E-5.
When NaOH is added to the acidic buffer solution, the following reaction occurs:
-----------(1)
Thus, the concentration of acid in the buffer decreases and the that of the salt increases.
Total volume of the solution after adding NaOH
= (433.5 + 433.5 + 6.12) mL
= 873.12 mL
Initial moles of acid in the solution (number of moles = Molarity(mol/L) X Volume (L))
Initial moles of salt in the solution
Moles of NaOH added in the solution:
ICE table for equation (1)
We have final number of moles of the salt NaA and the acid HA. Their final concentration in the solution
= number of moles / Final total volume
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