pH of 0.2 M salt AB of weak acid HA and weak base BOH is (pK_a of HA = 4 and pK_b of BOH = 6)
(a) 4
(b) 6
(c) 7
(d) 9

For a salt of weak acid and weak base [pK_a- pK_b] would be equal to –

pH of a salt solution of wak acid (pK_(a) = 4) & weak base (pK_(b) = 5) at 25^(@)C is :

A weak acid HA has a pH =4 . This can be confirmed as

pH of 0.05 aqueous solution of weak acid HA (K_a=4×10^(-4)) is

The percentage degree of hydrolysis of a salt of weak acid (HA) and weak base (BOH) in its 0.1 M solution is found to be 10%. If the molarity of the solution is 0.05 M, the percentage hydrolysis of the salt should be :

What will the pH of the solution of the salt of weak acid and weak base? (K_(b) = 1 xx 10^(-6) and K_(a) = 1 xx 10^(-4) )

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 )

What is the pH of a 0.010 M solution of a weak acid HA that is 4.0 % ionized ?

pH of a solution of salt of weak acid and weak base is : pH=1/2pK_w+1/2pK_a-pK_b and that of weak acid and strong base is pH=1/2pK_w+1/2pK_a+1/2logc The pH of 0.1 M sodium acetate ( K_a for CH_3COOH=1.8xx10^(-5) ) is