Question
a buffer is made by adding 150 ml of .595M BaF2 and 0f .500M of HF solution.
calculate the pH of this buffer system
calculate the pH of this buffer after adding .100 mol Hcl
calculate the pH of this buffer after adding .0750 mol Ca(OH)2
ka(HF)= 6.9x10^-4
kb(F-)= 1.4 x 10^-11
calculate the pH of this buffer system
calculate the pH of this buffer after adding .100 mol Hcl
calculate the pH of this buffer after adding .0750 mol Ca(OH)2
ka(HF)= 6.9x10^-4
kb(F-)= 1.4 x 10^-11
Answers
Use the Henderson-Hasselbalch equation; BUT I don't think you can make a 0.6 M solution of BaF2. It isn't that soluble is it. Ksp is about 10^-6.
im confused on how to calculate the pH of this buffer after adding .100 mol Hcl ... i get negative numbers
You write the two equations for what goes on in the buffer solution.
F^- + H^+ ==> HF
So when you add acid (HCl), you decrease the F^- moles and you increase the HF moles. So for final base it is what you had initially - moles H^+ added. For final acid, it is what you had initially + moles H^+ added. Plug those new values into the log(base/acid) terms and recalculate. It won't change the pH very much.
F^- + H^+ ==> HF
So when you add acid (HCl), you decrease the F^- moles and you increase the HF moles. So for final base it is what you had initially - moles H^+ added. For final acid, it is what you had initially + moles H^+ added. Plug those new values into the log(base/acid) terms and recalculate. It won't change the pH very much.
i got ph= 3.24+log -7
b/c i did
ph=3.24 = log .175/-.025
b/c i did
ph=3.24 = log .175/-.025
It almost 24 hours since we worked on this. If you still need it, repost at the top of the page and I'll try to go through it with you.
no thank you
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