Question: Determine the volume of .125M NaOH required to titrate to the equivalence point of 25mL of a .175M of a weak acid that is 20% ionized
Answer is 35mL correct me if I'm wrong!!! HELP
Answer is 35mL correct me if I'm wrong!!! HELP
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20 = (x / 0.175)(100)
x = [H^+] = 3.5x10^-2 M
n(H^+) = (3.5x10^-2 M)(0.025 L) = 8.75x10^-4 mol <------ amount that reacts with NaOH
Volume = 8.75x10^-5 mol / 0.125 M = 7.0x10^-3 L = 7 mL
How did you get 35 mL?
x = [H^+] = 3.5x10^-2 M
n(H^+) = (3.5x10^-2 M)(0.025 L) = 8.75x10^-4 mol <------ amount that reacts with NaOH
Volume = 8.75x10^-5 mol / 0.125 M = 7.0x10^-3 L = 7 mL
How did you get 35 mL?