Question

A buffer solution was made by adding `15.0 g` of `CH_(3) COOH` and `20.5gCH_(3)COONa`. The buffer is diluted to `1.0L`.
a. Calculate the `pH` of solution.
b. What will be the change in `pH` if `10.0mL` of `1.0 M HC1` is added to it.
Given: `pK_(a) of CH_(3)COOH = 4.74, log ((13)/(12)) = 0.035`

Answer 1

Correct Answer - A::B::C::D
i. Mw of `CH_(3)COOH = 60 g mol^(-1)`
Mw of `CH_(3)COONa = 82g mol^(-1)`
`["Salt"] = (20.5)/(82) = 0.25M (V = 1L)`
`["Acid"] = (15.5)/(60) = 0.25M (V = 1L)`
`pH = pK_(a) + log.(["Salt"])/(["Acid"])`
`= 4.74 + log.([0.25])/([0.25]) = 4.74`
ii. Rule `A A A:` In acidic buffer `(A)` on adding of `S_(A)(A)`, the concentration of `W_(A)(A)` increases and that of salt decrease.
Moles of `HC1 = (10xx1M)/(1000) = 10^(-2)M = 0.01M`
`[Acid] = 0.25 + 0.01 = 0.26M`
`["Salt"] = 0.25 - 0.01 = 0.024`
`pH_("new") = pK_(a) + "log" [("Salt")/("Acid")] = pK_(a) + log ((0.24)/(0.26))`
`= pK_(a) + log ((12)/(13)) = pK_(a) - log ((13)/(12))`
`= 4.74 - 0.035 = 4.705`
Change in `pH:`
`DeltapH = pH_("new") - pH_("inital")`
`= 4.705 - 4.74 =- 0.035`