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Please find Ka for Question 3 and 4, I have left the work but my professor has said it isn't correct and left the comments I attatched about them. Please help! pH values should be correct though
pH vs. mLNaOH added
You have to factor in dilution at this point, whenever base volume is added, the concentration changes even before reaction. Also, note: when base is added it removes from the concentration of the acid and creates conjugate base. 4) same issues as (3)
3) Ka at 1/2 equivalence (or midpoint) of NaOH Equivalence point would be 17.0/2=8.5 mL, and pH=4.73 Ka=(0.170−x)x2​pH=−log[H+]→[H+]=10−pH=10−4.73=0.0000186=xKa=(0.170−0.0000186)0.00001862​=2.2∗10−8​
4) Ka at 10 mL added NaOH Ka=(0.175−x)x2​ pH is around 4.9 pH=−log[H+]→[H+]=10−pH=10−4.9=1.26∗10−5=x Ka=(0.00175−1.26∗10−5)0.00001262​=9.14∗10−8