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Expert Solution

first question

Let the equal volume be V so HNO2 and NaNO2 are taken V/2 volume each.

so [NaNO2]/[HNO2]=(0.30 mol/L * V/2)/(0.60 mol/L * V/2)=1/2

also pka=-log ka=-log (4.46 *10^-4)=4-0.65=3.35

now pH= pH=pka + log [base]/[acid]

             =pka + log [NO2-]/[HNO2]

          =3.35+ log 1/2=3.35-0.3010=3.049=3.05 option B

second question

Total volume of solution=1L

[acetic acid]=[Na acetate]=0.1mol/L

pH=pka + log [base]/[acid]

or, pH=pka + log [acetate]/[acetic acid]

        =4.74 + log 0.1M/0.1M , M=mol/L

       =4.74 +0

        =4.74

now when 0.01mol of NaoH is added.

NaOH + CH3COOH=CH3COOH+H2O

Equilibrium conc of acetic acid=0.1-0.01=0.09                                 acetate=0.1+0.09=1.222

So pH = pka + log [acetate]/[acetic acid]=4.84 + log 1.222=4.84 + 0.087

inc in pH=0.087 option C

queen_honey_blossom answered 2 hours ago
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