What is the pKb of a base whose conjugate acid has a pKa of 6.11?

Answer 1

Take it away from 14.0 to give 7.89.

Answer 2

so, any time you have questions like this, start up questioning Henderson-Hasselbalch: the equation kinds: pH = pKa + log([A-]/[HA}) or pH = pKa - log ([HA]/[A-]) Now, that's log base ten, so extremely, the linked fee of the logarithm is what ability do you ought to enhance the quantity ten to get the linked fee interior the parentheses. for area a, in view that HA and A- are at equivalent concentrations, the log is of a million. log of a million is comparable to 0. subsequently, your pH must be the pKa. thus, 6.5 for area b, the linked fee of [A-]/[HA] works out to be 10, so log of that could paintings out to be a million. subsequently, pH = pKa +a million or 7.5 for area C, [A-]/[HA] works out to be .a million, so log of that is -a million, so pH is pKa + (-a million) so, 5.5 area d is working backwards. pH = 4.5, which ability log([A-]/[HA]) ought to equivalent -2. So what's 10^(-2)? .001 with a bit of luck you are able to make certain something from here.