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## Main Question or Discussion Point

Here's the problem how much of 5M KOH should be added to 1L of 0.1M glycine(pKa=9.6) at a pH of 9 until the pH reaches exactly 10? the answer in the back says 10mL.

R-NH3(+) <-> R-NH2 + H(+)

Here's the way I solved it. I figured out the number of moles of R-NH2 using the henderson-hasselbach equation, and it is .251 when the pH was 9.

And then i figured out the increase in moles of R-NH2 when the pH is at 10:

pH = pKa + log(NH2+x / NH3(+) - x)

10 = 9.6 + log((moles of NH2+x)/(moles of NH3(+) - x))

x = .064

then for every mole of R-NH2 dissociated a mole of KOH must be used.

so concentration * volume = .064

5M * V = .064

V ~ 13 mL

This is not the correct answer, although it might seem close. How did they end up with 10 mL? I tried rounding it to 0.06 but that only gave me 12 mL. Am i being too precise here, and the book just rounded down to 10 mL due to significant figures? Please help. Thanks

R-NH3(+) <-> R-NH2 + H(+)

Here's the way I solved it. I figured out the number of moles of R-NH2 using the henderson-hasselbach equation, and it is .251 when the pH was 9.

And then i figured out the increase in moles of R-NH2 when the pH is at 10:

pH = pKa + log(NH2+x / NH3(+) - x)

10 = 9.6 + log((moles of NH2+x)/(moles of NH3(+) - x))

x = .064

then for every mole of R-NH2 dissociated a mole of KOH must be used.

so concentration * volume = .064

5M * V = .064

V ~ 13 mL

This is not the correct answer, although it might seem close. How did they end up with 10 mL? I tried rounding it to 0.06 but that only gave me 12 mL. Am i being too precise here, and the book just rounded down to 10 mL due to significant figures? Please help. Thanks