Question

Human blood contains a buffer of carbonic acid (H ₂CO ₃) and bicarbonate anion (HCO ₃^-) in order to maintain blood pH between 7.35 and 7.45, as a value higher than 7.8 or lower than 6.8 can lead to death.

Carbonic acid is about 0.108 percent and the density is 1.67 g/mL. Determine the molarity of carbonic acid in blood.

Answer 1

CONCEPT OF MOLARITY Given:- Per cent w/w of carbonic acid = 0.108 % Density of carbonic acid = 1.67 g-ml" Molecular weight of carbonic acid (H2CO3) = 2+12+48 = 62 g-mol Relation between molarity (M) and percent of concentration:- Weight/weight % (w/w %) - weight of solute (g) weight of solution (9) x 100 ........(i) weight of solute in (9) Molarity (M) = molecular weight of solute (g mol-1) Х 1000 volume of solution in ml (ii) Density of solution (d) weight of solution volume of solution Rewrite as, volume of solution = weight of solution density of solution (iii) From equation (ii) and (iii), value of equation (iii) put in equation (ii), we get, weight of solute in (g) 1000 Molarity (M) molecular weight of solute (g mol-1) weight of solution in garm density of solution in gram -ml-1 1000 Molarity (M) weight of solute in (g) xdensity of solution (g ml-?) molecular weight of solute (g mol-1) weight of solution in garm Rewrite equation weight of solute in (g) xdensity of solution (g ml-1) Molarity (M) = weight of solution in garm 1000 molecular weight of solute (g mol-1 Again, rewrite equation weight of solute in (g) Molarity (M) = weight of solution in garm X 100 X 10 X density of solution (g ml-1 molecular weight of solute (g mol-1

On comparing with equation (i), we get Final equation M= w/w % x10xd M.wt. (iv) Now apply equation (iv), M= 0.108 % x 10 x 1.67 gml-1 0.0291 mol L-1 = 0.030 mol L-1 62 gmol-1 Answer :- Molarity of carbonic acid (H2CO3) in is 0.030 mol-L-1