pH calculation Flashcards
You work in a chemistry lab, and are asked to prepared 250 mL of a H2PO4-/HPO42- buffer in which the concentration of the weak acid component is 0.0316 M and the concentration of the conjugate base is 0.0565 M. Your starting materials are 0.527 M H2PO4-, 1.00 M NaOH, and distilled water. What volume (in mL) of the acid (the 0.527 M H2PO4-) do you need to prepare this buffer solution (answer in three sigfigs)
Since the starting materials are water, H2PO4-, and NaOH, the reaction would be (for the sake of this question, the arrow denotes equilibrium, not resonance!!)
H2PO4- + OH- ⇔HPO42- + H2O
From the question, molarity and volume of H2PO4- and its conjugate base are given. So H2PO4- has 0.007900 mol and HPO42- has 0.01413 mol in buffer solution.
Since the starting material of H2PO4- contributes both 0.007900 mol H2PO4- and 0.01413 mol HPO42-, 0.007900 mol + 0.01413 mol = 0.02203 mol of starting material H2PO4- is needed to make the buffer.
0.01413 mol of NaOH is also needed to make the conjugate base.
Therefore, 0.02203 mol / 0.527 M of starting reagent of weak acid * 1000 mL = 41.8 mL of H2PO4-
0.01413 mol / 1 M of NaOH * 1000 mL = 14.1 mL NaOH
Find the pH of a solution prepared by dissolving all of the following compounds in one beaker and diluting to a volume of 0.250 L: 0.100 mol benzoic acid (pKa = 4.20), 0.100 mol sodium benzoate, 0.020 mol H2SO4, and 0.050 mol NaOH. Assume sulfuric acid dissociatescompletely (i.e. it is a strong acid).
We need to recognize that the strong acid and strong base will react until one is consumed:
H2SO4 + 2 NaOH → Na2SO4 + 2 H2O
Start mol 0.020 0.050 0 0
End mol 0 0.010 0.020 0.04 0
Now, the excess strong acid will react:
HA + OH- → A- + H2O
Start mol 0.100 0.010 0.100 --
End mol 0.090 0 0.110 --
So, we now have a buffer with 0.090 mol benzoic acid and 0.110 moles benzoate:
pH = pKa + log mol A- = 4.20 + log 0.110 mol = 4.29
mol HA 0.090 mol
In the end, pH = 4.29
You are asked to prepare 500. mL of a 0.300 M acetate buffer at pH 5.10 using only pure acetic acid (MW=60.05 g/mol, pKa=4.76), 3.00 M NaOH, and water. What mass of acetic acid and volume of 3.00 M NaOH is required to make this buffer?
First, determine the ratio of A-/HA needed for pH 5.10 by rearranging the Henderson Hasslebach equation:
(mol A-)/(mol HA) = 10pH-pKa = 105.10-4.76 = 2.1878
We also know that the total moles of HA and A- = 0.500 L * 0.300M = 0.1500 mol.
So:
mol A- + mol HA = 0.1500 and
mol A- = 2.1878 mol HA
Solving for each term, we find: mol HA = 0.04705 mol and mol A- = 0.10295 mol
Quantities for each:
We need a total of 0.1500 moles of acetic acid:
0.1500 mol HA * 60.05 g HA = 9.01 g acetic acid
1 mol HA
We need to convert 0.10295 moles of the acetic acid to acetate by adding NaOH
0.10295 mol HA * (1 mol NaOH/1 mol HA) * (1000 mL/3 mol NaOH) = **34.3 mL ** 3M NaOH
- Your new employer has asked you to prepare 1.00 L of a pH 12.00 buffer with a total
phosphate concentration of 0.0500 M. You have at your disposal the following compounds
Compound Ka Molar Mass (g/mol)
H3PO4 7.11 x 10-3 97.9950
NaH2PO4 6.34 x 10-8 119.9769
Na2HPO4 4.22 x 10-13 141.9588
Na3PO4 – 163.9407
Which two compounds would you use to prepare a buffer of pH 12.00 and how many grams of each of the two selected compounds would you need? (12 points)
When preparing a buffer, it is best to choose and acid/base pair whose pKa is as close to the target pH as possible. This optimizes buffer capacity. In the case of the items at our disposal, the Na2HPO4/ Na3PO4 combination seems best since the pKa for HPO42- is 12.37.
First, determine the ratio of PO43-/HPO42- needed for pH 12.00 by rearranging the Henderson Hasslebach equation.: mol PO43- = 10pH-pKa = 1012-12.37 = 0.422 mol HPO42-
We also know that the total moles of PO43- and HPO42- = 1L x 0.0500M =0.0500 mol.
So:
mol PO43- + mol HPO42- = 0.0500 and mol PO43- = 0.422 mol HPO42-
Solving for each term, we find: mol PO43- = 0.0148 mol and mol HPO42- = 0.0352 mol
Masses for each salt:
0.0148 mol Na3PO4 x 163.9407 g Na3PO4 = 2.43 g Na3PO4
1 mol Na3PO4
0.0352mol Na2HPO4 x 141.9588 g Na2HPO4 = 4.99 g Na2HPO4
1 mol Na2HPO4
How would you prepare 10mL of a 0.01M phosphate buffer, pH 7.40, from stock solutions of 0.10M KH2PO4 and 0.25M K2HPO4? pKa of KH2PO4 = 7.20?
1. Use the Henderson Hasselbalch equation to find the ratio of A- to HA.
pH = pKa + log [A-] / [HA]
7.40 = 7.20 + log [A-] / [HA]
0.20 = log [A-] / [HA]
1.584893192 = [A-] / [HA]*
*Since [A-] / [HA] = 1.584893192, we can say that [A-] / [HA] = 1.584893192/ 1. In this case [A-] = 1.584893192; [HA] = 1.
2. Calculate the decimal fraction (part/whole) of each buffer component.
A- = 1.584893192 / (1.000 + 1.584893192) = 1.584893192 / 2.584893192= 0.61313682
HA = 1.000 / 2.584893192= 0.38686318
3. Find the molarity (M) of each component in the buffer by simply multiplying the molarity of the buffer by the decimal fraction of each component.
MA- = 0.01M * 0.61313682 = 0.006131368M
MHA = 0.01M * 0.38686318 = 0.003868632M
4. Calculate the moles of each component in the buffer.
Moles = Molarity \* Liters of buffer molesA<sup>-</sup> = 0.006131368M \* 0.01L = 6.131 x 10-5 moles
molesHA = 0.003868632M x 0.01L = 3.869 x 10-5 moles
5. Calculate the volume of each stock solution required to make the buffer
Liters of stock = moles of the buffer component / Molarity of the stock
LA- = 6.131 x 10-5 moles / 0.25 M = 2.452 x 10-4 L = 245μL
LHA = 3.869 x 10-5 moles / 0.10 M = 3.869 x 10-4 L = 387μL
6. To prepare this buffer, one would use appropriately-sized pipets to measure and transfer each component to a 10mL volumetric flask and bring the solution to volume with dH2O.
How would you prepare 10mL of a 0.02M acetate buffer, pH 4.30, from stock solutions of 0.05M acetic acid (HAc) and 0.05M NaOH? pKa acetic acid = 4.76?
1. Use the Henderson Hasselbalch equation to find the ratio of A- to HA.
pH = pKa + log [A-] / [HA]
4.30 = 4.76 + log [A-] / [HA]
-0.46 = log [A-] / [HA]
0.34673685 = [A-] / [HA]
2. Calculate the decimal fraction (part/whole) of each buffer component.
A- = 0.34673685 / (1.00 + 0.34673685) = 0.34673685 / 1.34673685 = 0.257464441
HA = 1.00 / 1.34673685 = 0.742535559
3. Find the molarity (M) of each component in the buffer by simply multiplying the molarity of the buffer by the decimal fraction of each component.
MA- = 0.02M x 0. 257464441= 0.005149289 = 5.15 x 10-3 M
MHA = 0.02M x 0. 742535559= 0.014850711 = 1.49 x 10-2 M
4. Calculate the moles of each component in the buffer. Moles = Molarity x Liters of buffer
molesA- = 0.005149289M x 0.01L = 5.14929 x 10-5 moles
molesHA = 0.014850711M x 0.01L = 1.485071.47 x 10-4 moles
5. Note: Since this buffer is prepared by the reaction of a weak acid (HAc) with a strong base (NaOH), you must determine the total moles of the weak acid component needed because the conjugate base is made in situ.
Total moles = 5.15 x 10-5moles NaOH + 1.49 x 10-4 moles HAc = 2.00 x 10-4 moles HAc. This sum indicates that, although in the buffer one only needs 1.49 x 10-4 moles HA, an additional 5.15 x 10-5moles is needed to generate the conjugate base in situ.
6. Calculate the volume of each stock solution required to make the bufferLiters of stock = Moles of the buffer component / Molarity of the stock
LA- = 5.14929 x 10-5 moles / 0.05 M = 1.03 x 10-3L NaOH = 1.0mL
LHA = 2.00 x 10-4 moles / 0.05 M = 4.00 x 10-3 L CH3COOH = 4.0mL
7. To prepare this buffer, one would use appropriately sized pipets or cylinders to measure and transfer each component to a 10 mL volumetric flask and bring the solution to volume with dH2O.
What is the pH of a solution resulting from a mixture of 200 mL of 0.1 M NaOH and 100 mL of 0.3 M HAc?
1. Calculate the initial moles of each reactant. Moles = Molarity x Liters
MolesNaOH = 0.1M x 0.2L = 0.02moles
MolesHAc = 0.3M x 0.1L = 0.03moles
2. Write the balanced equation for this reaction. Note: This is a neutralization reaction.
NaOH + HAc → Na+Ac- + H+OH-
3. Determine the moles of each reactant remaining after the reaction. Recall: This is a limiting reagent type problem. Since NaOH is present in the least amount and, according to the balanced equation the reactants react in a 1:1 ratio, the NaOH is completely consumed (i.e., zero moles of NaOH remain). Because the reactants react in a 1:1 ratio,
Moles HAc remaining = MolesHAc initial – MolesHAc reacted
= 0.03moles – 0.02moles = 0.01moles
4. Determine the molarity of the buffer components (i.e., the conjugate acid-base pair).
MHAc = molesHAc remaining / Liters solution = 0.01moles/0.3L = 0.033M
The molarity of the conjugate base Ac- resulting from the dissolution of the product NaAc is:
MAc- = moles formed / Liters solution = 0.02moles / 0.3L = 0.067M
5. Use the H-H equation to calculate the pH.
pH = pKa + log [A-] / [HA]
= 4.76 + log (0.067/0.033)
= 4.76 + log 2.02
= 4.76 + 0.31
= 5.07
Consider a solution of formic acid (HCHO2) and sodium formate (NaCHO2), at pH=3.70. After the addition of 0.015 moles of HNO3, the pH decreases by 0.12. What are the initial molarities of HCHO2 and CHO2-, if the Ka of HCHO2 is 1.8x10-4?
What is the pH of a buffer containing 0.20 M H2CO3 and 0.10 M NaHCO3?
(Ka of H2CO3=4.3*10-7)
weak acid equilibrium:
H2CO<u>3</u> + H2O ⇆ H3O+ + HCO3-
Ka= [H+][HCO3-]/[H2CO3]
isolate [H+] and plug in values.
Ans: 6.07
What ration of NaF to HF isnecessary to make a pH 3.90 buffer.
Ka of HF: 3.5*10-4
HF + H2O ⇆ F- + H3O+
Ka=[F-][H+]/[HF]
Derive to get: pH = pKa + log ([F-]/[HF])
plug in values to get ratio.
Ans: 2.8
You want to make a pH = 9.00 NH3/NH4+ buffer. How many grams of NH4Cl(s) do you need to add to 1.0 L of the 0.10M NH3 solution to make this buffer. (Assume the volume doesn’t change upon additon of the NH4Cl.) MW of NH4Cl = 53.49 g/mol
Kb of NH3=1.8*10-5
NH3 + H2O ⇆ NH4+ + OH-
Kb=[OH-][NH4+]/[OH-}
Kb[NH3]/[OH] = [NH4+]
Since pOH=14-pH = 14-9 = 5
[OH-] = 10-pOH = 10-5 M
1.8*10-5 (0.10 M)/(10-5 M) = 0.18 M NH4+
So: (0.18 M NH4+)*(1.0 L)*53.49 g/mol = 9.6 g
What weight of NaOAc·3H2O should be added to one liter of 0.100 M HOAc to prepare a
buffer solution having a pH of 5.24?
molar mass of NaOAc·3H2O = 136.08 g/mol
HOAc + H2O ⇆ H3O+ + OAc-
Ka=[-OAc][H3O+]/[HOAc]
Ka[HOAc]/[H3O+]=[-OAc]=1.8*10-5*0.100M/(1.8197*10-5M)
[OAc-] = 0.0989 M
So: 1L * 0.0989 M OAc- = 0.0989 mol OAc-
So: 0.0989 mol * 136.08 g/mol = 43 g
What weight of NaOH should be added to 1 liter of 1. 00 M HOAc to prepare a buffer
solution having a pH of 4.00?
6.1 g
