In first the concept we used was that the molarity of solution
is equal to the concentration of H+ and we solved for
the pH. In second question the formula which we used is that
concentration of H+ is equal to square root of ka.c where c is
concentration of weak acid.
Given, 10ml 20 me 012 0.1 M HNO₂ 0.2 M H₂SO4 a. - of O om a mixture of two strong acids for mixture of two different acid [at] = M = M,V, +. M. Va md e por tot = V tv + Vwater given question mixture is diluted to 100 mlie. Veot 2 100 ml O 0.1810 +0.2x20 - loo [ht] 5 20.05 M 100 pH = -log [H] = -109 ( 5 x109 pH = 1:301
Симеи OL All of the given acids are weal acids. so it concentration for given a cids is 647] = ke . cis concentration, pH = -log (Ska C) - - log (koc) 1 (109 ka + 109 loka - loge for 1.0M Phenol given, kaz1.3x10-10 DM 2-1109 ( 1.3x101 - pH = 4.943 b) for 0.08 m pyruvic acid pka - 2.76 pH e 2.76 - log (8x102)). 12.76+1.097) 0.834 5 1.9284
for olm Bromo-acetic acid pka = 4.75 I for pH = [(pka - log c) - I (4.75 - log 10') > ] (575) , 1.875 o.olm chloro-nitrous acid ko s 4.54103 pH -f log (4.8x10%x102) - Ixlog (4.5X1009) = 2:173 So; By comparison, the pH of lom phenol is least acidic and 0.00 m pyruvic aci - is most acidia