Grade 12 chemistry! Acid-Base Equilibrium?
1. Which of the following 1.0M salt solutions is acidic? A. BaS B.NH4Cl C.Ca(NO3)2 D.NaCH3COO 2. Water has the greatest tendency to act as a base with which of the following? A. HF B. H2CO3 C. H3PO4 D. CH3COOH 3. the H3O+ is greater than the OH in A.HCL B.NH3 C.H20 D.CH3COONa 4. a) HNO2 B) HCLO 3)HBr which of the above weak acids would form a buffer solution when its conjugate base is added? A. a) only B. b) only C. a) and b) only D, All
Chemistry 12 acids and bases help?
Thanks in advance! The following equilibrium favors the products: Se2 + HSo4 → Hse + So4 a) Which is the stronger of the two acids in the equation? b) If NaHSe were added to the solution, how would the equilibrium be affected?
Basic Chemistry Equilibrium Problems? Help?
I have a few equilibrium problems that I need help with. Even if you only answer one, it's fine. Please explain how you arrived at your answer if you can! If the pressure on the reaction N2 (g) + O2 (g) <==> 2NO (g) at equilibrium is increased, (Note: the "2"s after "N" and "O" are subscripted) a) The quantity of N2 (g) decreases. b) The quantity of NO (g) increases. c) The quantity of NO (g) decreases. d) The quantities in the system do not change. If the temperature of the equilibrium system X + Y <==> XY + 25 kJ decreases, a) [X] decreases and [XY] increases. b) [X] increases and [XY] decreases. c) [X] decreases and [XY] decreases. d) The concentrations of reactants and products do not change. Thanks much in advance! Hopefully these didn't strain your brain too much.
Help with Initial Change Equilibrium? Grade 12 Chemistry?
A) Quinine (C20H24N2O2) will dissolve into QH+ and OH- (Not really sure the exact equation) So lets set up the table ........Quinine.........QH+.......OH Initial.....1.7e-2........0..........0 Change......-x........+x.........+x Eq........1.7e-2-x......+x........+x Kb = [Conjugate Acid] * [OH-] / [Base] 3.3e-6 = x *x / 1.7e-2 - x (we can forget about the x here, since its so small, it won't affect the equation) 3.3e-6 = x *x / 1.7e-2 5.61e-8 = x^2 Now square root both side 2.37e-4 = x = [OH-] pOH = -log[2.37e-4] pOH = 3.62 pH = 14 - pOH pH = 14 - 3.62 pH = 10.37 Now for part B methanoic acidr is: HCOOH in water the equation is HCOOH <---> HCOO- + H+ so first we need the concentration of the acid. Concentration = Molarity = mols / liter (or 1000mL) So its already in concentration form = 0.40 M So now we need the ICE table .....HCOOH.......HCOO-..... H+ I.....0.40............0..............0 C....-x..............+x............+x Eq....0.40-x.........+x........+x Ka = [Conjugate Anion] * [H+] / [Acid] 1.8e-4 = x * x / 0.40-x 1.8e-4= x^2 / 0.40 7.2e-5 = x^2 0.00848 = x = [H+] pH = -log[H+] pH = -log[0.00848] pH = 2.07
Chemistry 12 acids/bases question?
Well, using the equation above. If you add heat to the equation it shifts to the left producing more H3O+ ions so the pH goes down. the pOH would therefore go up pKw would get larger as the amount of products increase in concentration.
Acids and Bases? word problem for chemistry?
the formula for [H3O+] is: [H3O+] = 10^(-pH) so, it will be 10^(-5.2) = 6.31x10^-6 HCN is acidic because: almost all of solutions that has Hydrogen at the front is acidic and... pH that is lower than 7 is acidic. on the other hand, pH greater than 7 is basic.
Grade 12 Chemistry Question. Weak Acids?
What is the concentration of a monoprotic weak acid if its pH is 5.50 and its Ka = 5.7 x 10^-10? This is how I would do it: [H+] = 10^-5.50 = .0000032 mol/L Ka= [H+][A-]/ [HA] 5.7 x 10^-10 = (0.0000032)^2/ [HA] [HA] = (0.0000032)^2/ 5.7 x 10-10 = 0.018 mol/L Therefore the concentration of the weak acid is 0.018 mol/L Can anyone confirm this is right or show me how to do it? Thank you
Chemistry Grade 12 pH Problem?
amount base added=025L Total volume=.070L n(HC2H3O2), before neutralization=.0045 n(OH-) added=.00575 n(C2H3O2-) formed .0045 n(OH-) excess .00125 From The excess OH- beyond the needed to neutralize all the acetic acid , we know that M(OH-)=.00125 mol/.00.070L=.017857 mol/L pOH=-log[OH-]=-log(.017857)=1.75 and pH=14-1.75=12.25