20)
HNO2 +NaOH ------------------> NaNO2 + H2O
20x0.1= 2 0 20x0.1=2 ----------- initial mmoles
---------- 10x0.05 = 0.5 -------------- ----------- change
1.5 0 2.5 --------- equilibrium
pKa of HNO2 = -log 7.1x10-4 = 3.15
The pH of this buffer = pKa + log [conjugate base]/[acid]
= 3.15 + log 2.5/1.5
= 3.37
21)
NH4Cl + NaOH ------------------> NH3 + NaCl + H2O
250x0.36 =90 0 250x0.3= 75 ----------------------- initial mmoles
----------- 50x0.5= 25 ---------- ----------------------- change
65 0 100 ------------ equilibrium
The pKb of NH3 = -log 1.8 x10-5 = 4.74
pOH of the base is given by Hendersen equation as
pOH = pKb + log [conjugate acid]/[base]
= 4.74 + log 65/100 = 4.55
and pH = 14-pOH = 14-4.55=9.45
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