TBR Acids and Bases Discrete #99

[MD2BeWpg]

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Q: What is the pH of 0.050 M H3CCH2CO2H (pKa for HCCH2CO2H = 4.89)

I understand that you use the formula: pH=1/2pKa-1/2log[HA].
The final equation is 2.45+0.65 = 3.10
I understand all of the math EXCEPT, why do you end up with +0.65 instead of -0.65?
In other words Why do you end up adding 0.65 to 2.45 instead of subtracting 0.65 from 2.45?

I realize this is a simple misunderstanding with my logs. Someone please help!

 

[loveoforganic]

pH = 1/2pKa - 1/2log[HA]

pH = 1/2(4.89) - (1/2)log(0.050)

pH = 2.45 - (1/2)(-1.3) : You end up with -1.3 here, because 10^(-number) is actually 1/(10^+number), a fraction

pH = 2.45 - (-0.65)

pH = 2.45 + 0.65

pH = 3.1

 

[MD2BeWpg]

pH = 1/2pKa - 1/2log[HA]

pH = 1/2(4.89) - (1/2)log(0.050)

pH = 2.45 - (1/2)(-1.3) : You end up with -1.3 here, because 10^(-number) is actually 1/(10^+number), a fraction

pH = 2.45 - (-0.65)

pH = 2.45 + 0.65

pH = 3.1

Thanks!