Calculate the [h3o+] and [oh-], pH and the percent dissociation for both 0.25 mol/L HCl and 0.25 mol/L HCN.
0.25 mol/L HCL
[H+] = 0.25 mol/L
[OH-] = 1 x 10^-14 / 0.25 = 4 x 10^-14
pH = 0.60
% dissociation = ?
( I don't know how to figure out percent dissociation)
For the HCN I thought I would do the same thing since there is also 0.25 mol/L of it but in the solution there is the number 1.2 x 10^ -5 and I don't understand where it came from. Could someone please help? thanks :)
% diss = 100% for HCl because there is no undissociated HCl present.
My guess is that 1.2 x 10^-5 is (H3O^+) but you can work it out.
Ka for HCN in my tables (but you use your tables) 6.2 x 10^-10. So set up an ICE chart and calculate (H3O^+) (which I find to be 1.2 x 10^-5), then
you can do pH, pOH, H^+, OH^-, and
% diss = [(H3O^+)/(HCN)]*100
Do we always have to do an ICE table for weak acids and bases?
Unless you have a lot of experience and can do some of it in your head, leave out some steps, and keep things straight. The ICE chart is too easy to use not to use it.
Did you get your KHP/Ca(OH)2 problem solved? I figures the problem to be you had not listed the mass KHP and you were calculating M Ca(OH)2.
Yes I figured out that problem. :) Thanks a lot!
To calculate the percent dissociation for an acid, we need to determine the initial concentration of the acid and its dissociation constant. The percent dissociation is the ratio of the concentration of dissociated ions to the initial concentration of the acid, multiplied by 100.
For the HCl solution:
[H+] = 0.25 mol/L (given)
[OH-] can be calculated using the equation [H+][OH-] = 1 x 10^-14 at 25°C (from the autoionization of water). Rearrange the equation to solve for [OH-]: [OH-] = 1 x 10^-14 / [H+]
Substituting the value of [H+] = 0.25 mol/L, we can calculate [OH-]:
[OH-] = 1 x 10^-14 / 0.25 = 4 x 10^-14 mol/L
pH is calculated as the negative logarithm (base 10) of the hydrogen ion concentration: pH = -log[H+]. Substituting the value of [H+] = 0.25 mol/L, we can calculate pH:
pH = -log(0.25) ≈ 0.60
To calculate the percent dissociation, we need additional information. The given concentration of HCl is 0.25 mol/L, but we need to know the dissociation constant (Ka) to determine the percent dissociation. Please provide the Ka value for HCl.
Regarding HCN:
The value of 1.2 x 10^-5 you mentioned is the dissociation constant (Ka) for HCN. It represents the equilibrium constant for the dissociation of HCN into H+ and CN- ions. This is a different situation than HCl, where HCl is a strong acid that completely ionizes in water. Hence, for HCN, we cannot use the simple 1:1 ratio as in HCl, and we need to consider the Ka value.
Once we have the Ka value for HCN, we can use it to determine the percent dissociation following a similar approach as for HCl. Please provide the Ka value for HCN so that we can continue with the calculations.