Calculate the concentrations of the protonated and unprotonated forms of bromothymol blue in 100 mL of a pH 8.?

Hi ,



pKa = 7,10 . . . (bromothymol blue)

pH = pKa + log([Base] / [Acid] )

log([Base] / [Acid] ) = pH - pKa
log([Base] / [Acid] ) = 8,1 - 7,10
log([Base] / [Acid] ) = 1,0

[Base] / [Acid] = 10

n(bromothymol blue) = C x V
n(bromothymol blue) = 0,005 x 1?10^-3
n(bromothymol blue) = 5?10^-6 mol


[protonated] / [unprotonated] = 10

[protonated] = 10 x [unprotonated]


n(protonated) = 10 x (unprotonated)

n(protonated) + n(unprotonated) = 5?10^-6 mol

n(protonated) = 10 x (unprotonated)


n(protonated) = 4,545?10^-7 mol

n(unprotonated) = 4,545?10^-6 mol



[protonated] = n(protonated) / 0,100
[protonated] = 4,545?10^-7 / 0,100
[protonated] = 4,545?10^-6 mol/L


[unprotonated] = n(unprotonated) / 0,100
[unprotonated] = 4,545?10^-6 / 0,100
[unprotonated] = 4,545?10^-5 mol/L