Question

I don't know how to approach this problem.

How many mL of a concentrated HNO3 does a student have to dilute to 2.60 L in order to prepare a solution with a pH of 1.67? Conc HNO3, has the following information on its label: 47% HNO3, by mass, Den = 1.38 g/ml, MW HNO3 = 63 g/mol

[H+] = 0.02

[OH-] = 4.08 x 10^-13

Answer 1

47% by mass

D = 1.38 g /ml

MW = 63 g/mol

first, calculate molarity of acid

M = mol/V

assume a basis of

100 g of solution

so

%of HNO3 = 0.47*100 = 47 g of HNO3 are present

that is

mol = mass/MW = 47/63 = 0.7460 mol of HNO3

we need volume of solution so

D = M/V

V = M/D = 100/1.38 = 72.4637 ml

now, change to liter = 72.4637/1000 = 0.0724637

calculat emolarity

M = mol/V = 0.7460 /0.0724637 = 10.29480 M of HNO3

now.. calculate pH = 1.67

[H+] = 10^-pH = 10^-1.67 = 0.02137

assume

diluted [HNO3] = [H+] = 0.02137

V1 = ?

M1 = 10.29480

V2 = 2.6

M1 = 0.02137

apply dilution law

M1V1 = M2V2

10.29480 *V1 = 2.6*0.02137

V1 = 2.6*0.02137/10.29480 = 0.0053970 L

V1 = 5.39 ml