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A solution prepared by mixing 10 mL of 1 M HCl and 10 mL of 1.2 M NaOH has a ph of
18. Aqueous Equilibrium
Titrations: Strong Acid-Strong Base
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A 50.0 mL sample of 0.200 M sodium hydroxide is titrated with 0.200 M nitric acid. Calculate the pH in the titration after the addition of 60.0 mL of 0.200 M HNO3.
A
1.00
B
2.00
C
7.00
D
12.00
Verified step by step guidance1
Determine the initial moles of NaOH using the formula: \( \text{moles} = \text{concentration} \times \text{volume} \). For NaOH, \( \text{moles} = 0.200 \, \text{M} \times 0.0500 \, \text{L} \).
Calculate the moles of HNO3 added using the same formula: \( \text{moles} = 0.200 \, \text{M} \times 0.0600 \, \text{L} \).
Determine the limiting reactant by comparing the moles of NaOH and HNO3. Since both have the same concentration and volume, they will react in a 1:1 ratio.
Calculate the moles of excess HNO3 or NaOH after the reaction. Subtract the smaller number of moles from the larger to find the excess.
If HNO3 is in excess, calculate the concentration of \( \text{H}^+ \) ions in the solution using the formula: \( \text{[H}^+\text{]} = \frac{\text{moles of excess HNO3}}{\text{total volume in L}} \). Then, find the pH using \( \text{pH} = -\log(\text{[H}^+\text{]}) \).
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