How many moles of nitric acid do you need to prepare \(200 \mathrm{~mL}\) of an aqueous solution that has a pH of \(2.0\) ?

Short Answer

Expert verified
To prepare \(200 \mathrm{~mL}\) of an aqueous solution with a pH of 2.0, you'll need \(0.002\) moles of nitric acid (HNO₃).

Step by step solution

01

Calculate [H⁺] from pH

Using the pH formula, we can find the concentration of H⁺ ions in the solution. The formula to find [H⁺] is: \[ [H⁺] = 10^{-pH} \] Given the pH value as 2.0, plug in the value into the equation: \[ [H⁺] = 10^{-2} \]
02

Determine the concentration of HNO₃

Since nitric acid, HNO₃, is a strong acid, it will dissociate completely in water, producing equal concentrations of H⁺ and NO₃⁻ ions. The dissociation of nitric acid is as follows: \[ HNO₃ (aq) \rightarrow H⁺ (aq) + NO₃⁻ (aq) \] Therefore, the concentration of HNO₃ in the solution is equal to the concentration of H⁺ ions, which we calculated in Step 1.
03

Calculate the moles of HNO₃ required

Now, we can find the moles of HNO₃ needed for the 200 mL aqueous solution. We will apply the formula: \[moles= Molarity * Volume \] First, let's convert the volume from mL to L: \[Volume (L) = \frac{Volume (mL)}{1000} = \frac{200}{1000} = 0.2 L\] Now plug the values into the equation: \[moles = [H⁺] * Volume \] \[moles = 10^{-2} * 0.2 \]
04

Calculate the final answer

Multiply the values from Step 3: \[moles = 0.01 * 0.2 = 0.002 \] So, to prepare 200 mL of an aqueous solution with a pH of 2.0, 0.002 moles of nitric acid (HNO₃) are needed.

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