Unisciti a migliaia di studenti che si affidano a noi per superare al meglio i loro esami!
Scelta multipla
What is the calculated pH after adding 2.5 mL of 0.5 M NaOH to 50.0 mL of 0.25 M acetic acid, and then adding 1.0 mL of 0.5 M HCl?
A
3.50
B
7.00
C
4.76
D
5.00
0 Commenti
Guida verificata passo dopo passo
1
Calculate the moles of NaOH added: Use the formula \( \text{moles} = \text{volume (L)} \times \text{concentration (M)} \). Convert 2.5 mL to liters and multiply by 0.5 M to find the moles of NaOH.
Calculate the moles of acetic acid initially present: Convert 50.0 mL to liters and multiply by 0.25 M to find the moles of acetic acid.
Determine the moles of HCl added: Convert 1.0 mL to liters and multiply by 0.5 M to find the moles of HCl.
Calculate the moles of acetic acid and acetate after the reaction with NaOH: Subtract the moles of NaOH from the moles of acetic acid to find the remaining moles of acetic acid. The moles of acetate formed will be equal to the moles of NaOH added.
Calculate the pH using the Henderson-Hasselbalch equation: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{acetate}]}{[\text{acetic acid}]} \right) \). Use the moles of acetate and acetic acid to find their concentrations and substitute into the equation, using the \( \text{pK}_a \) of acetic acid (4.76).