Due to highway accident, 150 L of concentrated hydrochloric acid (12.0 M) is rel
ID: 529878 • Letter: D
Question
Due to highway accident, 150 L of concentrated hydrochloric acid (12.0 M) is released into a lake containing 5.0 times 10^L of water, if the pH of this lake was 7.0 prior to the accident, what is the pH of the lake following the accident? Calculate the pH of a buffer solution prepared by dissolving 0.2 mole of sodium cyanate (NaCNO) and 1.0 mole of cyanic acid (HCNO) in enough water to make 1.0 liter of solution (K_g(HCNO) = 2.0 times 10^-) Calculate the molar solubility of CaF_2 in a 0.25 M solution of NaF(aq)_2 (K_sp (CaF_2) = 4.0 times 10^11) Calculate Delta G degree for the reaction 3NO_2(g) + H_2O(l) rightarrow 2HNO_2(l) + NO(g) Delta G degree_f (kJ/mol) H_2O(l) -237.2 HNO_3(l) - 79.9 NO(g) 86.7 NO_2(g) 51.8 Hydrogen peroxide (H_2O_2) decomposes according to the equation H_2O_2(l) rightarrow H_2O(l) + 1/2 O_2(g) Calculate K_p for this reaction at 25 degree C. (Delta H degree = -98.2 k/mol, Delta S degree = 70.1 J/K middot mol Given the following cell diagram. Al(s) | Al^3+ (aq) || Cd^2+ (aq) | Cd(s) a) what is the balanced overall (net) cell reaction? b) Calculate E degree for the cell Calculate the cell emf for the following at 25 degree C Ni(s) + 2Cu^2+ (0.010 M) rightarrow Ni^2+ (0.0010 M) + 2 Cu^+ (1.0 M) For the electrochemical cell, Cd(s) | Cd^2+ (aq) || Co^2+ (aq) | Co(s), determine the equilibrium (K_eq) at 25 degree C for the reaction that occurs. A current of 0.80 A was applied to an electrolytic cell containing molten CdCl_2 for 2.5 hour Calculate the mass of cadmium metal deposited.Explanation / Answer
22)
Step -1Calculate moles of HCl –
n = c V
c = molar concentration (the unit is moles/liter)
V = volume (the unit is liters)
Moles HCl = 12.0 mol/L x 150L = 1800 moles
Step-2 Calculate [H+]
Volume of water in lake = 5.0 x 10^8 L
[H+] = 1800 mol/ 5.0 x 10^8L =3.6 x 10^-6 M
Step-3 Calculate PH -
PH = - log [H+]
New pH = - log [3.6 x 10^-6] = 5.44
The pH of the lake following the accident = 5.44
23)
Concentration HCNO = 0.20 / 1.0 = 0.20 M
Concentration CNO- = 1.0 / 1.0 = 1.0 M
Ka (HCNO) =2.0 x 10^-4
PKa = - log Ka
PKa = - log 2.0 x 10^-4
PKa = 3.7
Now calculate PH by using Henderson–HasselBalch equation
PH = PKa + log [H-]/[HA]
PH = PKa + log [CNO-] / [HCNO]
PH = 3.7 + log 1.0 / 0.20
PH = 3.7 + 0.70 = 4.40