Question

how many moles are in 6ml of 85% H3PO4? and how many grams of sodium carbonate are needed to neutralize this acid?

Answer 1

(i)

100 gm of 85 wt% solution
mass of H3PO4, W = 85gm
density of 85 wt% solution, rho = 1.7 gm/ml
mass of solution, Ws = 100gm
volume of solution, V = Ws/rho = 100/1.7 = 58.8 ml = 0.0588 L

Molar mass H3PO4 = 3+31+4*16=98 gm/mol
mass of H3PO4, W = 85gm
no of moles of H3PO4, n = 85/98=0.867mol
Concentration in Molarity, C = n/V = 0.867/0.0588 =14.7 M

6ml of 85 wt% H3PO4
Volume of solution, Vs = 0.006 L
No of moles, n = C*Vs = 14.7*0.006 = 0.0882 mols

(ii)
Neutralization reaction
2H3PO4(aq) + 3Na2CO3(aq) → 3 H2O(l) + 3 CO2(g) + 2Na3PO4(aq)

Volume of solution, Vs = 0.006 L
nfactor of H3PO4 = 3 as it contain 3 H+ attached to Oxygen
Normality of H3PO4, N1 = Molarity*nfactor = 14.7*3 = 44 N

Let the normality concentration of Na2CO3 = N2 N
Balance of equivalence in neutralization reaction
N1*Vs = N2*Vs
So, N2 = N1 = 44 N

nfactor of Na2CO3 = 2 as it contains total +2 charge by cations
Molarity of Na2CO3, C2 = N2/nfactor = 44/2 = 22 M
Moles of Na2CO3, n2 = C2*Vs = 22*0.006 = 0.132 mol
Molar mass of Na2CO3, MW = 106 gm/mol
Mass of Na2CO3, W = n2*MW = 0.132*106 = 14gm