Abuffer is prepared by dissolving 0.80 moles of nh3 and 0.80 moles of nh4cl in 1.00 l of aqueous solution. if 0.10 mol of naoh is added to 250 ml of this buffer, what is the ph of the resultant solution?
a.10.54
b.4.27
c.8.78
d.9.73
e.5.22

a. 10.54

Explanation:

reaction is

NH₃ +  NaOH   NH₄Cl + H₂O

0.10 mol NaOH will consume 0.10 mol NH₃ thereby decreasing the initial amount of moles NH₃ and increasing that of NH₄Cl

mol NH₃  = 0.80 - 010 = .70

mol NH₄Cl = 0.80 + .10 = 0.90

pH = pKₐ + log (( NH₃/NH₄Cl))

pH =  9.26 + log (( 0.90 + 0.70)) = 9.26 + 0.11 = 10.54

1 a

2 d

3 a

4 c

explanation:

wher is the rest of the question