First slide

PH of strong acids and strong bases

Question

The pH of a buffer solution is 4.745. When 0.044 mole of Ba(OH)₂ is added to 1 lit. of the buffer, the pH changes to 4.756. Then the buffer capacity is

Easy

A

4
B

0.25
C

0.5
D

8

Solution

0.044 mol of Ba(OH)₂ = 0.044 x 2 mol of OH^-
When 0.088 mol of OH^- is added, change in P^H
= (4.756 - 4.745)
= 0.011
Conversely,
0.011 change of P^H occurs when moles of OH^- added = 0.088 mol
1 change of P^H occurs when moles of OH^- added $= \frac{0.088}{0.011} \times 1 = 8 mol$

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