Please answer questions b-e b. How many individual hydroxide ions (OHf\') are fo
ID: 580888 • Letter: P
Question
Please answer questions b-e b. How many individual hydroxide ions (OHf') are found in 134 ml of 0 586 M BaOH(a? c. What volume (in L) of 0 586 M BatOH1) (ay) contains 0.466 ounces of BarOf), dissolved in i? d. If 16.0 mL of water are added to 31.5 mL of 0.586 M Ba(OH) (ay what is the new solution molarity? e. Suppose you had titrated your vinegar sample with barium bydroside instead of sodium bydroxide What volume (in ml) of 0.586 M Ba(OH (ay) must be added to a 5.00 ml sample of equivalence point? Use your average vinegar molarity (see page 1) in this calculation vinegar to reach theExplanation / Answer
b) Consider the ionization of Ba(OH)2 as below.
Ba(OH)2 (aq) -------> Ba2+ (aq) + 2 OH- (aq)
As per the stoichiometric equation,
1 mole Ba(OH)2 = 2 moles OH-.
Moles of Ba(OH)2 in 13.4 mL of 0.586 M Ba(OH)2 = (13.4 mL)*(1 L/1000 mL)*(0.586 M)*(1 mol L-1/1 M) = 0.0078524 mole (1 M = 1 mol/L).
Moles of OH- = (0.0078524 mole Ba(OH)2)*(2 mole OH-/1 mole Ba(OH)2) = 0.0157048 mole.
As per Avogadro, 1 mole of any substance contains 6.02*1023 molecules/ion of the substance; therefore, number of ions of OH- corresponding to 0.0157048 mole OH- = (0.0157048 mole OH-)*(6.02*1023 OH- ions/1 mole OH-) = 9.45429*1021 OH- ions 9.454*1021 OH- ions (ans).
c) We shall use the relation 1 international ounce = 28.35 g.
The solution contains 0.466 ounce Ba(OH)2; therefore, the weight of Ba(OH)2 in the solution = (0.466 ounce)*(28.35 g/1 ounce) = 13.2111 g.
Molar mass of Ba(OH)2 = (1*137.327 + 2*15.9994 + 2*1.008) g/mol = 171.3418 g/mol.
Mole(s) of Ba(OH)2 corresponding to 13.2111 g Ba(OH)2 = (13.2111 g)/(171.3418 g/mol) = 0.07710 mole.
Volume of Ba(OH)2 required in L = (moles of Ba(OH)2)/(molarity of Ba(OH)2) = (0.07710 mole)/[(0.586 M)*(1 mol L-1/1 M)] = 0.1315699 L 0.1316 L (ans).
d) We shall use the dilution equation, M1*V1 = M2*V2 where
M1 = 0.586 M, V1 = 31.5 mL and V2 = (31.5 + 16.0) mL = 47.5 mL.
Plug in values and get
(0.586 M)*(31.5 mL) = M2*(47.5 mL)
====> M2 = (0.586*31.5)/(47.5) M = 0.388610 M 0.3886 M.
The new molarity of the Ba(OH)2 solution is 0.3886 M (ans).
e) I need to know the molarity of vinegar to answer the question.