Asked by Always
Should 25 ml of 0.15 M HCL solution be added to 250 mL of buffer solution containing 0.25M HF Ka-7.1 x 10, and 0.40 M NaF at a pH of 3.15 what is the new pH?
Answers
Answered by
DrBob222
millimols NaF = mL x M = approx 100
mmols HF = mL x M = aprox 62.5
mmols HCl added = approx 3.75
You omitted the exponent on Ka. I think that's 7.1E-4, then pKa = 3.14
................F^- + H^+ ==> HF
I..............100....0.......62.5
add.................3.75
C...........-3.75..-3.75......+3.75
E............96.25....0.......66.25
Substitute the E line into HH equation and solve for pH. On the surface it sounds like much too much HCl is added; however, the pH shows there is enough buffer capacity there to handle it (and more).
mmols HF = mL x M = aprox 62.5
mmols HCl added = approx 3.75
You omitted the exponent on Ka. I think that's 7.1E-4, then pKa = 3.14
................F^- + H^+ ==> HF
I..............100....0.......62.5
add.................3.75
C...........-3.75..-3.75......+3.75
E............96.25....0.......66.25
Substitute the E line into HH equation and solve for pH. On the surface it sounds like much too much HCl is added; however, the pH shows there is enough buffer capacity there to handle it (and more).
Answered by
Always
For the formula pH=pka + log(base/acid)
Which would be the base and which would be the acid? Thanks.
Which would be the base and which would be the acid? Thanks.
Answered by
DrBob222
NaF is the base
HF is the acid.
HF is the acid.
Answered by
Always
Thank you so much!
Answered by
Always
Would HCL be put on the bottom with HF because it's an acid also?
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