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What is the pH of a 0.02 M NaOH solution?
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17. Acid and Base Equilibrium
pH of Strong Acids and Bases
Scelta multipla
What is the pH of the solution formed when 50 mL of 0.250 M NaOH is added to 50 mL of 0.120 M HCl?
A
2.00
B
7.00
C
12.00
D
1.00
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Calculate the moles of NaOH using the formula: \( \text{moles} = \text{concentration} \times \text{volume} \). For NaOH, use 0.250 M and 50 mL (convert mL to L by dividing by 1000).
Calculate the moles of HCl using the same formula: \( \text{moles} = \text{concentration} \times \text{volume} \). For HCl, use 0.120 M and 50 mL (convert mL to L by dividing by 1000).
Determine the limiting reactant by comparing the moles of NaOH and HCl. The reactant with fewer moles is the limiting reactant.
Calculate the moles of excess NaOH remaining after the reaction by subtracting the moles of HCl from the moles of NaOH.
Determine the concentration of the excess NaOH in the final solution by dividing the moles of excess NaOH by the total volume of the solution (100 mL, converted to L). Then, calculate the pH using the formula: \( \text{pH} = 14 - \text{pOH} \), where \( \text{pOH} = -\log[\text{OH}^-] \).
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