0) A sample of concentration nitric acid has a density of 1.41 g/ml and contain

Question

0) A sample of concentration nitric acid has a density of 1.41 g/ml and contain 70% 10) HNO3 by mass, Show all calculation set ups. answer the question beloq: (a) What mass (g) of HNO3 is present per liter of solution? (density Mass/volume (b) What is the molarity of the HNO3 solution? (d) In the lab, the acid spill is usually neutralized by using NaHCO3 as shown below HNO3(aq) NaHCO3(s) NaNO3(aq)H200) CO2(8) How many gram of NaHCO3 is required to neutralized 3.0 ml of the original concentrated nitriic acid?

Explanation / Answer

70% HNO3 by mass
That is 70 g of HNO3 in 100 g of the solution.

Density = 1.41 g/mL

So, mass = Density x volume
                = 1.41 g/mL x 1000 mL
               = 1410 g

Now, 70% of 1410 g = (70/100) x 1410 g = 987 g

(b)

Mass of HNO3 = 987 g
Molar mass of HNO3 = 63 g/mol

So, moles of HNO3 = (987 g) / (63 g/mol) = 15.67 moles

Molarity = moles / Liter
               = 15.67 moles / 1 L
             = 15.67 M

(d)

HNO3(aq)   +   NaHCO3(s)     ------->      NaNO3(aqq)   +   H2O(l)   +   CO2(g)

The reaction equation is already balanced.

In the above reaction equation HNO3 and NaHCO3 reacts in 1:1 ratio.

Now,

3 mL of orginal solution.

Volume = 3 mL
density = 1.41 g/mL
So, mass = 3 mL x 1.41 g/mL = 4.23 g

Since, the solution is 70% HNO3 by mass.

So, mass of HNO3 present in 3 mL of original solution = 70% of 4.23 g
                                                                                                = (70/100) x 4.23 g
                                                                                                = 2.96 g

Moles of HNO3 = (2.96 g) / (63 g/mol) = 0.047 mol

So, moles of NaHCO3 = 0.047 mol

Molar mass of NaHCO3 = 84 g/mol

So, mass of HNO3 = (0.047 mol) x (84 g/mol) = 3.95 g

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