Question

Hello there. I need help with working on some calculations for two vials (with numbers 3 and 6). We did a lab that measured three vials. In vial #1, it had a pH of 9, 5 mL of solution was added, and the moles were 5x10^-8 mol OH-. The actual pH was 6.69 for vial #1.

1. Calculate the number of moles of either OH- or H3O+ in the solution that was added. (OH-) if the solution was a base or H3O+ if the solution was an acid) and calculate the expected pH of the solution in the vial

Equation: H3O+(aq)+OH-(aq) -> 2H2O(l)

In vial #3, I said it was a base (1.0x10^-11 aka pH of 11) and 1mL of solution was added.

and in vial #6 I said it was an acid (1.0x10^-1 aka pH of 1) and 1mL of solution was added.

I'm lost on how to calculate the moles and then the pH. Below is what I have so far and I would really appreciate some guidance. Thank you!

Data Analysis / Lap Report vial 1l measured pH/ calculated pH | % error 9.05 (6.69 35.3% 8.80 questions Net ionic equation HCl cast NaOH a H₂Ole)+ NaCl(aa) H+ (aq) + C1-(99) + Natt Olt (aq) → H₂O(l) + Nat(en) + C1-(aa) ho+ (aq) +OH(aq) >2H₂0 (1) 2) concentration in #1 = 1.0 x 10.9 Colt] = -log(1,0 x 10-9) = [OH] = 9 14-9=5 1.0x 105 x x= 5x10-6 mol OH. 0.005 3) Polt= -log (5 x 10 8 mol = 7.301 pH = 14-7.301 = 6.69 percent error = (4.05. - 6.69) = 0.353 → 35.31. 6.69

Answer 1

PAGE NO. DATE / imi - soli Vial3 calculation for Vial 3 Griver, base is added ph of base solution is 11 we know that pu= -log[ht ] where, Cht is concentration of Hiicns. Pont where, -log[on] con] is concentration of onions. na Putpont 14 - Puna 14-pu SC14- Pon=3 Pon= 3 = = Con - -log[on] 103 mole Con number of moles of on wolume of solution in L n conta [on] = [on] 10TL

103 L In 156 mole su, number of moles of our is 1100 mole Calculation for vial of Ginn, and is added Phy, acid solution is t we know, Phi -ing [nt] 12 - Ing[nt] [htJa is [nt] is concentrates a ntion [nl] = no. of moles antics Avolume of solution 43. n S - ych) in [at nimi 1034 n = 154 mul ant So, number of mules of Htion is 109 mole

from number a moles Calculation of pn ph= lug[nt] [nt] - w.qmules of nties Volume of solubic) er vial - - no. cfmules of mine 104 Volume y sol= Imolosh [nt] = 154 - 16 ph= -log[151] = 1 for vial 3 no. of moles a ona lob mole volume of on = 1032 [on] = ww.y moles of on- Volume q soluta [on]- 155 - Conta 163 Pon= -log[on] Poy = -log[103] on 3 Pn= luopun a 14-3 Pn all