Consider the following statements
I.The pH

Options

A. I and II
B. I, II and III
C. II and III
D. I, II and IV

Answer

B

Explanation

I.` underset((excess))underset(40mmol)(H_(2)SO_(4)) to underset(("Limit i n g rea g ent"))underset(40mmol)( Na_(2)SO_(4)) + 2H_(2)O`
After reaction 20 mmol 0 20 mmol 40 mmol
Total volume =400+400=800mmol
Number of moles of `H_92)SO_(4)=20mmol`
Number of moles of `H^(+)` ion =`2xx20m` mol=40m mol`
[H^(+)]=40/800=0.05`
`pH=-log[H^(+)]=-log(0.05)=2-log(5)=2-0.7=1.3`
pH=1.3
II.`K_(w)=[H^(+)][OH^(-)]`
As T increases, `K_(w)` also increases
III.pH=5 and `K_(a)=10^(-5)`
`implies-log[H^(+)]=5`
Weak monobasic acid ‘HA’ having concentration ‘C’
`HA(aq) to H^(+)(aq) + A^(-1)(aq)`
At t=0 C 0 0
At equilibirium `C-Calpha Calpha Calpha`
`K_(a)=([H^(+)][A^(-1)])/([HA])=(calpha)/(C(1-alpha))`
`alpha to `degree of dissociation
`therefore K_(a)=(Calpha^(2))/(1-alpha)=10^(-5) ` … (1) (given)
`[H^+)]=Calpha=10^(-5)`… (2)
From equation (1) & (2)
`alpha/(1-alpha)=1impliesalpha=1-alpha implies alpha=1/2`
`% alpha=1/2 xx100=50%`
IV. It is applicable