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?
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).
For the formula pH=pka + log(base/acid)
Which would be the base and which would be the acid? Thanks.
NaF is the base
HF is the acid.
Thank you so much!
Would HCL be put on the bottom with HF because it's an acid also?
To determine the new pH after adding the HCl solution, we need to consider the reaction that occurs between HCl and the components of the buffer solution. In this case, the reaction will take place between HCl and the weak base, HF (hydrofluoric acid), to form the conjugate acid, F- (fluoride ion), and water.
Given the volume and concentration of the HCl solution (25 mL and 0.15 M, respectively), we can calculate the moles of HCl using the formula:
moles of solute = volume of solution (L) x concentration (M)
First, convert the volume of the HCl solution from mL to L:
25 mL = 25/1000 = 0.025 L
Now, calculate the moles of HCl:
moles HCl = 0.025 L x 0.15 M = 0.00375 moles
Since the molar ratio between HCl and HF is 1:1, the moles of HCl will react with an equal amount of moles of HF, resulting in the formation of the same amount of moles of F-.
Now, let's calculate the molarity of HF and F- in the buffer solution. Since the volumes of the buffer components are not given, we assume that the volumes are additive:
Volume of HF = 250 mL + 25 mL = 275 mL = 0.275 L
Volume of NaF = 250 mL + 25 mL = 275 mL = 0.275 L
moles of HF = Volume (L) x Concentration (M)
moles of HF = 0.275 L x 0.25 M = 0.06875 moles
moles of F- = Volume (L) x Concentration (M)
moles of F- = 0.275 L x 0.40 M = 0.11 moles
Since the moles of HCl are less than the moles of HF in the buffer solution, the moles of F- formed will equal the moles of HCl added.
Now, let's calculate the concentration of F- in the buffer solution after the reaction has occurred:
New concentration of F- = moles of F- formed / total volume of the buffer solution
Total volume of buffer solution = volume of HF + volume of NaF
Total volume of buffer solution = 0.275 L + 0.275 L = 0.55 L
New concentration of F- = 0.00375 moles / 0.55 L ≈ 0.00682 M
Finally, to calculate the pH of the buffer solution after the addition of HCl, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
where [A-]/[HA] represents the ratio of the concentration of the conjugate base (F-) to the weak acid (HF).
pH = pKa + log (0.00682 M / 0.25 M)
Given that the pKa for HF is 7.1 x 10^-7, we can substitute the values and calculate the new pH:
pH = -log(7.1 x 10^-7) + log (0.00682 M / 0.25 M)
pH ≈ -(-6.15) + (-1.6)
pH ≈ 6.15 - 1.6
pH ≈ 4.55
Therefore, the new pH of the buffer solution after adding 25 mL of 0.15 M HCl is approximately 4.55.