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What is the pH of a solution of 1.00 L of water with 0.0387 moles of HNO₃ in it?

User Zvoase
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1 Answer

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Final answer:

The pH of a solution with 0.0387 moles of HNO3 in 1.00 L of water is approximately 1.412. To calculate this, we use the molarity of HNO3, which is 0.0387 M due to its complete dissociation, and then apply the pH formula, which is -log[H+].

Step-by-step explanation:

To find the pH of a solution with 0.0387 moles of HNO₃ dissolved in 1.00 L of water, we first need to calculate the molarity of HNO₃. Since molarity is the number of moles of solute per liter of solution, the molarity of HNO₃ here is 0.0387 M. Nitric acid (HNO₃) is a strong acid, meaning it dissociates completely in water to give one mole of H+ ions per mole of nitric acid. Therefore, the hydrogen ion concentration [H+] in the solution is also 0.0387 M.

Now we can use the formula for pH, which is pH = -log[H+]. Plugging in the hydrogen ion concentration:

pH = -log[0.0387]
= -(-1.412) (using a calculator)
= 1.412

The pH of the solution is approximately 1.412.

User Cphill
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