In the titration of 58.00mL of 0.500 M lactic acid (Ka= 1.4x10^-4) with 0.250 M potassium hydroxide, what is the pH after 58.00mL of the base has been added?
Added by Juan C.
Step 1
- Volume of lactic acid = 58.00 mL = 0.05800 L - Concentration of lactic acid = 0.500 M - Moles of lactic acid = Volume × Concentration = 0.05800 L × 0.500 mol/L = 0.02900 moles Show more…
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