Question

What is the PH of a 100 mM solution of sulfuric acid (H2SO4)?

Ka1=Large

Ka2=1.2x10^-2

A. 2.02

B. -0.96

C. 1.00

D. 1.9

E. 0.96

Answer 1

[H2504 ] = 100mm = 0.10 .. (1000 mm = 1) completely means during H₂SO4 dimoclates first disociation Kai is large it in Hi and HSO4 & first disocration: H2SO4 (aql After dwociation = [ht] = 0.10M Ht (aq) + HSoy- (aq) 0.10 M 0.10 M1 Kq₂ = 1.2x10-2 is small. It means Hooy do not dissociate completely while remain in equilibrium with ions during second disociation and Ht from first dissociation exerts tommon ion effect on dissociation of Hou- - second dissociation: lei (CE Table concentration (M) HSO4 (99) Ht (aq) + soy? 0.10 0,10 - C tx + x 0.10-x 0.10tx Kaa= [H+] [5042-] 1.2810 -2 = (0.1otx) x [Hoy ] 10.10-x) x² +0.112x - 0.0012 =0 Solving this quadratic equation , we get value of x and x cannot have negative value for ar2+bx+c=0 Here, a=1 x= - b ± √b2490 b= 0.112 2a Co -0.0012

x= -0.112 + (0.112)2 - 4X 1 X (-0.0012) x = 9,85X10-3M [ht] =(0.10 + x) = (0.10+ 9.85 × 10-3) M = 0.10985 M = 0.96 pH = -log[ht] I pH = 0.96 = -log (0.10985) Answer E is correct . Hence, option Please rate