The pH of `1 M PO_(4)^(3-) (aq)` solution is,
[ Given `pK_(b)=(PO_(4)^(3-))=1.62]`

The pH of 1 M PO_(4)^(3-) (aq) solution (given pK_(b) of PO_(4)^(3-) = 2 ) is

The pH of 1 M PO_(4)^(3-) (aq) solution (given pK_(b) of PO_(4)^(3-) = 2 ) is

When a salt reacts with water resulting into formation of acidic or basic solution, the process is referred to as salt hydrolysis. The pH of salt solution can be calculated using the following equations. pH = ( 1)/(2) ( p K_(w) + pK_(a) + log C) for salt of weak acid and strong base. pH = (1)/(2) ( pK_(w)- pK_(b) - log C ) for salt of weak base and strong acid. pH = (1)/(2) ( pK_(w) + pK_(a) - pK_(b)) for salt of weak acid and weak base The pH of 1M PO_(4(aq))^(3-) soluiton will be ( given pK_(b) of PO_(4)^(3-) = 1.62 )

Calculate pH of 1 M PO_4^-3) (aq) solution. pk_b(PO_4^-3) = 1.62

What is the normality of 1 M H_(3)PO_(4) solution ?

At what concentration of PO_(4)^(3-) ions in 0.1(M) aqueous solution of AgNO_(3), will Ag_(3)PO_(4) start to precipitate?Given: K_(sp)(Ag_(3)PO_(4))=1.3xx10^(-20) .

The normality of 1.5M H_(3)PO_(4) is-

The normality of 1.5M H_(3)PO_(4) is-

The normality of 1.5M H_(3)PO_(4) is-