Question

A 1.44 L buffer solution consists of 0.137 M butanoic acid and 0.275 M sodium butanoate. Calculate the pH of the solution following the addition of 0.069 moles of NaOH . Assume that any contribution of the NaOH to the volume of the solution is negligible. The Ka of butanoic acid is 1.52×10−5 .

Answer 1

The answer is "
`P^(H)=5.379`".

Step-by-step explanation:

`\ NaOH \ value = (n)/(v)\\\\`

`=(0.069\ moles)/(0.144L)\\\\=0.04791\ M`

`\ Ka=1.52 * 10^(-5)\\\\P^(ka) = -10g \ ka \\\\`

`= -10 * 1.52 * 10^(-5)\\\\= 4.82\\`

Equation:

`CH_3CH_2CH2COOH+NaOH\rightarrow CH_3CH_2CH_2COONa +H_2O\\\\`

`\boxed{\left\begin{array}{ccccc}I &0.137 &0.04791 &0.275 & -- \\ C &-0.04791 &-0.04791 &+0.04791 & -- \\E &0.08909 &0&0.32291 & -- \end{array}\right}`

`P^(H)= P^(ka)+\log(CH_3CH_2CH_2COONa)/(CH_3CH_2CH_2COOH)\\\\`

`= 4.82+\log(0.32291)/(0.08909)\\\\=5.379`