Calculate the amount of `NH_(4)C1` required to dissolve in `500mL` of water to have a `pH = 4.5, K_(b) = 2.0 xx 10^(-5)`.

Explanation

`[H^(+)]=10^(-pH)=10^(-4.5)=3.162xx10^(-5)M`
`NH_(4)^(+)+H_(2)OhArrNH_(4)OH+H^(+)`
`C(1-h)" "Ch" "Ch`
`K_(h)=Ch^(2)=(K_(w))/(K_(b))=(10^(-14))/(1.8xx10^(-5))=5.5xx10^(-10)`
`h=(K_(h))/(Ch)=(K_(h))/([H^(+)])=(5.5xx10^(-10))/(3.162xx10^(-5))=174xx10^(-5)`
`C=([H^(+)])/(h)=(3.162xx10^(-5))/(1.74xx10^(-5))=1.8mol//L`
500 ml of `H_(2)O` contains `(1.8)/(2)=0.9` mole
Mass in gm `=0.9xx53.5=48.15g`