Question

[REMOVED]

Answer 1

8.When base is added to water, the ions float apart leading to extra OH^- ions in the water: MOH → OH^- + M⁺. The resulting large concentration of OH^- makes the solution more basic and leads to a dramatic increase in the pH. ( since the product of concentrations, [OH^-][H⁺], remains fixed at K_w=10^-14, as the concentration of OH^- ions goes up, the concentration of H⁺ ions goes down.)

9. Percent change of pH for water( taking values from the given table,pH at 5 drops=12.54 and pH at 0 drops= 7.00)

= 100×[(12.54-7.00)/7.00]

=79.14

10. Percent change of pH for buffer( taking values from the given table,pH at 5 drops= 7.24 and pH at 0 drops = 7.00)

= 100×[(7.24-7.00)/7.00]

=3.43

11.yes comparing the percent change in pH for 5 drops of added base both in case of water and buffer, we can see the value is very very less for buffer than water. There is no drastic change in pH of buffer like water so definately these percentages support a resistance to change for the buffer solution.

12. If a strong base is added to a buffer, the weak acid in the buffer will give up its H⁺ in order to transform the base (OH^-) into water (H₂O) and itself to the conjugate base: HA + OH^- → A^- + H₂O. Since the added OH^- is consumed by this reaction, the pH will change only slightly.