The antilog of x is 10^x, not 10^-x If you're confusing this with the antilog used to "undo" the pH equation, remember that pH = -log[H3O+], so when we use the antilog function in the pH equation, we need to include the negative sign from that equation.
Why did you do 10^0.333 and not 10^-0.333 with a negative sign?
The antilog of x is 10^x, not 10^-x
If you're confusing this with the antilog used to "undo" the pH equation, remember that pH = -log[H3O+], so when we use the antilog function in the pH equation, we need to include the negative sign from that equation.
How did you get 3.347?
pKa = -log(Ka) and the problem says the Ka = 4.5 x 10^-4
@@RoxiHulet Thank you, How would you find the mass if the Kb was given instead of the Ka
@@adammostafa2741 find the mass of what?