Calculate the `pH` of a solution of given mixture.
a. `(2g CH_(3)COOH +3g CH_(3) COONa)` in `100mL` of mixture.
b. `5mL` of `0.1M NH_(4)OH + 250mL of 0.1 M NH_(4)C1`.
c. `(0.25 "mol of" CH_(3)COOH + 0.35 "mol of" CH_(3) COONa)` in `500mL` mixture.
`K_(a) "of" CH_(3)COOH = 1.8 xx 10^(-5) (pK_(a) = 4.7447)`
`K_(b) "of" NH_(4)OH = 1.8 xx 10^(-5) (pK_(b) = 4.7447)`

Answer

A, B, C, D

Explanation

a. It is acidic buffer.
`["Salt"] = (3 xx 1000)/(82xx100) = 0.365M`
`["Acid"] = (2 xx 1000)/(60 xx 100) = 0.33M`
`pH = pK_(a) + log.(["Salt"])/(["Acid"])`
`=4.7447 + log.((0.365)/(0.33))`
`= 4.7447 log 1.1`
`= 4.7447 + 0.0414 = 4.7861`
b. Total volume `= 250 +5 = 255 mL`
mEq of salt `= 250 xx 0.1 = 25`
`mEq` of base `= 5 xx 0.1 = 0.5`
`["Salt"] = (25)/(255), [Base] = (0.5)/(255)`
`pOH = pK_(b) + log.(["Salt"])/(["Base"])`
`= 4.7447 + log((25//255)/(0.5//255))`
`= 4.7447 + log 50`
`= 4.7447 + 1.7 = 6.4447 = 6.44` (Take `log 5~~0.7`)
`pH = 14 - 6.44 = 7.56`
c. `pH = pK_(a) + log .[("Salt")/("Acid")]`
`= 4.7447 + log ((0.35//500)/(0.25//500))`
`= 4.7447 + log(1.4) = 4.7447 + 0.1461 = 4.89`