pH=pka + log([CH3COO-]/[CH3COOH])

[CH3COOH]=n/v = 0,25 mole/ 250*10^-3 L = 1 mole/L

[CH3COO-]= n/v = 0,8 mole / 250 *10^-3 L = 3,2 mole / L

so : log([CH3COO-]/[CH3COOH])= log(3,2/1) = 0,5051499783

So : pH= 5,2651

Good Luck

what did u plug into the calculater to find pH. Was it the inverse log 0.505?

log( 3,2 ) = 0,505 and the inverse is 10^(0,505) = 3,2

When you did the inverse did you add the negative? 10^(-0.505)?

i keep getting 0.3125

Secure Information

Content will be erased after question is completed.

Enter the email address associated with your account, and we will email you a link to reset your password.

Forgot your password?

Sign Up