Question

Nitrous acid was added to 100 gallons of pure water until the pH of the solution dropped to 3.50. How much acid (in grams) was added to the water? Relevant information can be found near the end of the Acid-Base PowerPoint presentation.

Answer #1

Sol:-

Given pH = 3.50

i.e - log [H^{+}] = 3.50

[because pH = - log [H^{+}] ]

so

[H^{+}] = 10^{-pH}

[H^{+}] = 10^{-3.50}

[H^{+}] =0.000316 M

we know

HNO_{2} is monobasic acid , therefore Molarity of
HNO_{2} i.e

[HNO_{2}] = [H^{+}] = 0.000316 M

also

Molarity = Number of moles of solute i.e HNO_{2} /
volume of solution in L

Number of moles of solute i.e HNO_{2} = mass of
HNO_{2} / gram molar mass of HNO_{2}

so

Molarity = mass of HNO_{2} / gram molar mass of
HNO_{2} x volume of solution in L
............**.(1)**

gram molar mass of HNO_{2} = 47.013 g/mol

Volume of solution = 100 gallons = 378.541 L

because 1 gallon = 3.78541 L

Now from equation **(1)**, we have

0.000316 M = Mass of HNO_{2} / (47.013 g/mol) (378.541 L
)

Mass of HNO_{2} = 0.000316 mol/L x 47.013 g/mol x
378.541 L

Mass of HNO_{2} = 5.62 g

**Hence Mass of HNO _{2} added is equal to 5.62
g**

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