Use the Henderson-Hasselbalch equation.
pH = pKa + log [(base)/(acid)]
pKa for NH3 is Kw/Kb. You were given Kb.
The base is NH3. The acid is NH4Cl (the NH4^+ in NH4Cl).
Post your work if you get stuck.
Consider ammonia, what is the pH of a buffer solution prepared by adding 15.0 g of ammonium chloride to 5.00L of 0.200 M ammonia? ( the K for NH3 is given to me 1.80 x 10 to the power of -5)
6 answers
what is the Kw?
Ion product for H2O. Mathematically it is
(H^+)(OH^-) = Kw = 1 x 10^-14 and pKw = 14
The problem doesn't gave a Ka for NH3 but it gives a Kb; therefore, KaKb = Kw and Ka = Kw/Kb = 1 x 10^-14/1.80 x 10^-5 = ??
(H^+)(OH^-) = Kw = 1 x 10^-14 and pKw = 14
The problem doesn't gave a Ka for NH3 but it gives a Kb; therefore, KaKb = Kw and Ka = Kw/Kb = 1 x 10^-14/1.80 x 10^-5 = ??
i found the pH to be 9.75, i don't know if i did it right,
Assuming i did it right, there is another question stating what is the pH of the buffer after 90.0mL of 2.00M HCL are added?...
Assuming i did it right, there is another question stating what is the pH of the buffer after 90.0mL of 2.00M HCL are added?...
I worked it and obtained 9.80 but I could have goofed. Here is what I did; you need to confirm it.
kb = 1.80 x 10^-5. pKb = 4.74 and that subtracted from pKw of 14 = 9.26
pH = 9.26 + (1.00/0.28)
pH = 9.26 + 3.57 = 9.26+0.55 = 9.81.
(NH4Cl) = 15/53.49 = 0.280 and in 1 L makes it 0.280 M.
For the next part, calculate the number of moles added with 90.0 mL of 2.00 M HCl. Buffers work (basic ones), when acids such as HCl are added, by neutralizing the strong acid with the NH3. The reaction HCl + NH3 ==>NH4Cl + H2O decreases the NH3 concn and increases the NH4^+ concn by the number of moles HCl added. Just recalculate the new concns of base and acid, plug into the H-H equation. and redetermine the pH. If you want it checked, please repost (with problem) at the top of the board and someone will look at it.
kb = 1.80 x 10^-5. pKb = 4.74 and that subtracted from pKw of 14 = 9.26
pH = 9.26 + (1.00/0.28)
pH = 9.26 + 3.57 = 9.26+0.55 = 9.81.
(NH4Cl) = 15/53.49 = 0.280 and in 1 L makes it 0.280 M.
For the next part, calculate the number of moles added with 90.0 mL of 2.00 M HCl. Buffers work (basic ones), when acids such as HCl are added, by neutralizing the strong acid with the NH3. The reaction HCl + NH3 ==>NH4Cl + H2O decreases the NH3 concn and increases the NH4^+ concn by the number of moles HCl added. Just recalculate the new concns of base and acid, plug into the H-H equation. and redetermine the pH. If you want it checked, please repost (with problem) at the top of the board and someone will look at it.
Can I just response to you without an account?
1 answer