Need help with b and d CHEM 1100 pll: Measurement and Uses ion. The One of the m
ID: 1043118 • Letter: N
Question
Need help with b and d CHEM 1100 pll: Measurement and Uses ion. The One of the most important properties of aqueous soluations is the concentration of hydrogen concentration of H (or H)?) affects the solubility af ?norganic and organic species, t complex metal cations and the rates of chemical reactions the nature of For convenience the concenstration of H is frequently expressed as the pll of the solation rather than hydrogen ion molarity, pht is defined by the equation pH-lgli 1(1) whee the logarithm is aken to the base 10 IrI is LOx to moles per lier,the plt of the solution is 4.0. If(H 1-5.0x 10 M, the pll is 1.3 Basic solutions are also desenibed in terms of pht. In water the following relation exists H ][OH]-K-1.0x10 at 25 'C ) Since [H J equals (OH ] in pure water, by Equation 2, H ] must be 1 x 10 M. Therefore, the pH of disilled water is 7. Solutions in which (u (OHr) are acidic and will have a phi 7. A solution with a pH of 10 will have [H)-1 10 Mand [OH]-I x 10 M pH Measurement p?? can be determined experimentally in two ways. One is to use an indicator, a soluble dye whose color is sensitive to p/t. Indicator colors change over a relatively short (about 2-unit) pH range. When properly chosen, they give the approximate pl of solutions. Two common indicators are litmus, usually used on paper, and phenolphthalein, commonly used in acid-base titrations. Litmus changes from red to blue in the pll range 6 to 8. Phenolphthalein changes from colorless to red in the 8 to 10. Any one indicator is usefiul for determining phH only in the region where it changes papers, indicators and change color over a wide pl range, are also in color. Indicators are available for measuring pH in any part of the scale. Universal indicator which contain a mixture of several common useExplanation / Answer
Hello,
The Part A of your question set uses the concept of Buffers. As you must be aware of the fact that Buffer solutions are the ones which resist change in pH when a very minute amount of acid or base is added to it.
Now this experiment involves preparation of an acidic buffer which employs a weak acid CH3COOH and it’s salt CH3COONa. This salt is formed when the weak acid reacts with a strong base (NaOH).
You might be interested to know that during a titration between a weak acid and strong base, a buffer solution gets formed right after you add the 1st drop of titrant (NaOH) to the beaker containing weak acid & continues until the end point (ideally just before the end point). In this case the moment the titrant (NaOH) is added to the titration flask containing HA, an acid base reaction takes place, some weak acid is utilized in the process of consuming all the titrant (acid) and an equivalent amount of salt is formed. The following reaction would thus be established in the titration flask
HA + NaOH = NaA + H2O
This solution turns out to behave as a buffer solution since some unreacted weak acid would be simultaneously be present with the conjugate base (salt) & the pH of such a solution is calculated using Henderson’s Equation which can be represented as:
pH = pKa + log10[salt/acid]
For a monoprotic acid [salt] = [conjugate base]
Salt (formed) or acid (unreacted) concentration is measured in moles.litre-1
Now we shall use this equation to solve the questions in Part A using the pH data and Henderson’s Equation. I am taking the liberty of solving all the questions in Part A as I can see that some answers have been written incorrectly.
pH of Buffer Solutions
Solution
Measured pH
Beaker 1, Buffer
4.41
Beaker 2, Buffer
4.42
Beaker 3, Distilled Water
8.85
Beaker 4, Distilled Water
9.06
Beaker 1, Buffer + HCl
3.89
Beaker 2, Buffer + NaOH
4.87
Beaker 3, Distilled water + HCl
1.00
Beaker 4, Distilled water + NaOH
13.06
a) what is the pH of the buffer solution you prepared?
Corresponds to pH of Beaker 1 i.e. 4.41
b) Using the Henderson’s equation, calculate the theoretical pH of the buffer solution you prepared. The Ka of CH3COOH is 1.8 x 10-5
It’s given that:
Weight of CH3COONa = 4.2 grams
Molecular weight of CH3COONa = 82 grams
Moles of CH3COONa = 4.2/82 = 0.0512 moles
Moles of CH3COOH = Molarity x Volume (in litres) = 6 M x (8.5 x 10-3) litres = 51 x 10-3 moles
Since it is the ratio of concentrations, so volume would get cancelled out and we can work out the answer using only the number of moles.
pKa of CH3COOH = -log10[1.8 x 10-5] = 4.74
Applying Henderson’s equation, we get
pH = pKa + log10[salt/acid]
or, pH = 4.74 + log10[0.0512/(51 x 10-3)]
or, pH = 4.741
c) What is the pH of buffer solution after you add HCl ?
As seen from the pH data, this corresponds to the pH of Beaker 1 after addition of HCl which is 3.89
d) Calculate the pH of the buffer solution after you add HCl?
Moles of HCl added = Molarity x Volume = 6 M x (1 x 10-3) Litres = 6 x 10-3 moles
Millimoles of HCl = 6 millimoles
When HCl is added, it consumes some amount of CH3COO- to form proportionate amount of CH3COOH
CH3COONa + HCl = CH3COOH + NaCl
Moles of CH3COONa = 0.0512 moles
Millimoles of CH3COONa = 51.2 millimoles
Moles of CH3COOH = 51 x 10-3 moles
Millimoles of CH3COOH = 51 millimoles
Millimoles of CH3COONa used up = 6 millimoles
Millimoles of CH3COONa left = 51.2 – 6 millimoles = 45.2 millimoles
Millimoles of CH3COOH formed = 6 millimoles
Total Millimoles of CH3COOH = 51 + 6 = 57 millimoles
Again, since the ratio of concentrations is employed in Henderson’s equation, so volume would get cancelled out and we can work out the answer using only the number of moles.
pH = pKa + log10[salt/acid]
or, pH = 4.74 + log10[45.2/57]
or, pH = 4.74 – 0.1 = 4.64
e) What is the pH of thw buffer after you add NaOH?
This corresponds to the pH of solution in beaker 2 to which NaOH is added in the buffer & the pH is 4.87
e) Calculate the pH of the buffer solution after you add NaOH?
Moles of NaOH added = Molarity x Volume = 6 M x (1 x 10-3) Litres = 6 x 10-3 moles
Millimoles of NaOH = 6 millimoles
When NaOH is added, it consumes some amount of CH3COOH to form proportionate amount of CH3COONa
CH3COOH + NaOH = CH3COONa + H2O
Moles of CH3COONa = 0.0512 moles
Millimoles of CH3COONa = 51.2 millimoles
Moles of CH3COOH = 51 x 10-3 moles
Millimoles of CH3COOH = 51 millimoles
Millimoles of CH3COOH used up = 6 millimoles
Millimoles of CH3COOH left = 51 – 6 millimoles = 45 millimoles
Millimoles of CH3COONa formed = 6 millimoles
Total Millimoles of CH3COONa = 51.2 + 6 = 57.2 millimoles
Again, since the ratio of concentrations is employed in Henderson’s equation, so volume would get cancelled out and we can work out the answer using only the number of moles.
pH = pKa + log10[salt/acid]
or, pH = 4.74 + log10[57.2/45]
or, pH = 4.74 + 0.1 = 4.84
Hope this helps. I request you to take some time to rate the answer & to drop in your valuable feedback.
Please do post your queries pertaining to the solution provided (if any)
Thanks
pH of Buffer Solutions
Solution
Measured pH
Beaker 1, Buffer
4.41
Beaker 2, Buffer
4.42
Beaker 3, Distilled Water
8.85
Beaker 4, Distilled Water
9.06
Beaker 1, Buffer + HCl
3.89
Beaker 2, Buffer + NaOH
4.87
Beaker 3, Distilled water + HCl
1.00
Beaker 4, Distilled water + NaOH
13.06