Calculate the volume of 0.1 M hydrochloric acid (HCl) required to neutralize 50.0 mL of 0.15 M sodium hydroxide (NaOH) solution, and determine the pH of the resulting solution.
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To calculate the volume of hydrochloric acid (HCl) required to neutralize the sodium hydroxide (NaOH) solution, we use the equation:
Moles of acid (HCl) = Moles of base (NaOH)
First, let's calculate the moles of NaOH:
Moles of NaOH = Molarity × Volume in liters
= 0.15 M × 50.0 mL
= 0.15 mol/L × 0.050 L
= 0.0075 mol
Since HCl and NaOH react in a 1:1 ratio, the moles of HCl required to neutralize NaOH is also 0.0075 mol.
Next, let's calculate the volume of 0.1 M HCl required:
Volume of HCl = Moles of HCl / Molarity
= 0.0075 mol / 0.1 M
= 0.075 L
= 75.0 mL
So, 75.0 mL of 0.1 M HCl is required to neutralize 50.0 mL of 0.15 M NaOH.
To determine the pH of the resulting solution, we need to consider the reaction between HCl and NaOH. HCl is a strong acid, and NaOH is a strong base. When they react, they form water (H2O) and a salt (NaCl). Since both reactants are strong, the reaction goes to completion, resulting in a neutral solution with a pH of 7.
Therefore, the pH of the resulting solution is 7.
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