Question

9. Heavy water, is water in which the usual hydrogen atom, 'H, has been replaced with the heavy isotope of hydrogen, "H. At 25°C, the pH of pure heavy water is 7.476. Determine the autoprotolysis constant for heavy water. 10. When 1.23x10 moles of substance X was used to prepare 250.0 mL of an aqueous solution, the resulting solution had a pH of 3.47. Using the data identify substance X from the following table. compound anilinium chloride (CH NH.CI) 4.601 benzoic acid (C.HCO,H) hypoiodous acid (HOT) 1-napthol (C2H, OH) nitrous acid (HNO3) 2-nitrobenzoic acid (CH.NO.CO,H) 14.202 10.64 9.416 3.15 2.185
please do #9 and #10. please show details on how you got the answer and why you got it. just detail stuff please!

Answer 1

10. The auto-ionization of water can be written as follows: Molarity of acid = number of moles / volume in L = 1.23*10^-3/0.250 =4.92*10^-3 H2O(1) H(aq) +OH(aq) Therefore; K = [H,0"][OH-]-[H* [OH-]= 1.0x10-14 =0.00492 The dissociation reaction of the HA Weak acid , with H* (aq) as product is as follows: HA(aq) =H" (aq) + A- (aq) and expression for ionization constant (K, )is as follows: The value of K is used to calculate the concentration of Ho* or on in the solution. The value of K, depends only on temperature. K,= Here for heavy water pH = 7.476 [H+]= 10^- pH (HA The is pH is 3.47 Therefore, the concentration of hydrogen ion of the solution: [H"]=10*** [H"]=10-947 = 3.39x104 = 10^-7.476 So, =3.34*10^-8 Pure water is neutral so [H+] = [OH-]= 3.34*10^-8 And pKw = [H+] [OH-] =3.34*10^-8*3.34*10^-8 =1.12*10^-15 [A-] = 3.39x10-4 Now calculate the acid ionization constant as folllows: Now calculate the pKa =- log Ka =- log 2.3*10^-5 [HA] (x)(x) (1.31) (3.39x104)(3.39x10*) (0.00492) K = 2.3x10 = 4.63 Therefore the compound is anilinium chloride