[" The "pH" of "1MPO_(4)^(-3)(AQ)" solution will be (given "],[pKb" of "PO_(4)^(-3)=1.62)]

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 )

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

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

The pH of 0.2 M aqueous solution of NH_4Cl will be (pK_b of NH_4OH = 4.74, log 2 = 0.3)

A 2L solution was prepared by mixing of NH_(3) and (NH_(4))_(2)SO_(4) . If pH of solution was found 8 and solution contains one mole of NH_(3) then moles of (NH_(4))_(2)SO_(4) was taken will be [pKb of NH_(3)=5] 2 4 5 1

A 2L solution was prepared by mixing of NH_(3) and (NH_(4))_(2)SO_(4) . If pH of solution was found 8 and solution contains one mole of NH_(3) then moles of (NH_(4))_(2)SO_(4) was taken will be [pKb of NH_(3)=5] 2 4 5 1