Question

please answer both questions

QUESTION 28 Which of the following acids HCIO2(aq) HClO(aq) HBrO2(aq) HBrO(aq) HIO(aq) is the strongest in aqueous solution? a HIO(aq) HCIO(aq) HBrO2(aq) HCIO2(aq) e. HBrO(aq)

Answer 1

29. Order of Acidity

HClO2 > HBrO2.> HIO2 HBrO > HIO

Here Cl is more electronegative. When H attached to more electronegative atom , will be more acidic .

Hence HClO2 is more acidic .

Given, Molarity of H2CO3 = 0.184 M carbonic acid is a diprotic acid. The First dissociation of carbonic acid can be shown as below H2CO3(aq + H20 (1) <-----> HCO3- (aq) + H3O + (aq) Initial 0.1840 Change - X + x + x Equilibrium (0.184-x) Given Kal of Carbonic = 4.20E-07 Weak acid dissociation constant Kal = (HCO3-)(H30+)/(H2CO3) 4.20E-07 = x*x (0.184-x) 4.20E-07 = x^2 = 0.184 x^2 = 4.20E-07 * 0.184 sox ^2 = 7.73E-08 x = 0.00028 assume x is small then (0.184-x) will be 0.184 5% rule x/ initial con x 100 = 0.00028 = 0.151 % H3O+ (aq) 0.00028 0.184 x 100 < 5% Second dissociation of carbonic acid Ka2 is HCO3- (aq) + H20 (1) <-----> CO3 2- (aq) + Initial 0.00028 Change - y + y Equilibrium (0.00019-y) + y (0.00019+y) | Given Ka2= 4.80E-11 Weak acid dissociation constant Ka2= (CO3 2-)(H3O+)/(HCO3-) 4.80E-11 = y (0.00019+y) 4.80E-11 = y * 0.00028 4.80E-11 = y * 1.00 so 4.80E-11 = y = 4.800E-11 (0.00019-y) 0.00028 | 4.80E-11 = 2.78E-04 Total [H3O+]= (0.00019+y) = pH = log [H3O+] = -log[ 2.78E-04 0.00028 ] = + 3.56