icals Use Identity: Aor B : number:1 bleahote ng Identity of weak acid: phosbha+

Question

icals Use Identity: Aor B : number:1 bleahote ng Identity of weak acid: phosbha+ acid: phos phote nn m experimental data literature value NaOH] used Mass acid used of NaOH at endpoint .4mL 0 054 L cular weight of acid t NaOH ular weight of acid 134 5 pK, of acidCe om) 1.942291gmol ns for Member A: Calculate the pKa of your monoprotic acid using the pH observed when approximately 25% of the acid originally present was neutralized. Hint: what is the [base]/[acid] ratio at 25% neutralization? What is the volume of NaOH added at 25% neutralization? Calculate the expected initial pH of your monoprotic acid solution, using the literature pKa value and the actual concentration of your solution. Hint: what is the concentration of 2 g of your unknown acid in 50 mlL of water? Is this a solution of pure weak acid, pure weak base, or a mixture of the two? Calculate the pK, of your monoprotic acid using thepll observed srosiaey Calculate the pK, of your monoprotic acid using the pH observed when approximately 75% of the acid originally present was neutralized. Hint: what is the [base]/[acid] ratio at 75% neutralization? What is the volurne of NaOH added at 75% neutralization? Calculate the expected equivalence point pH of your monoprotic acid solution, using the literature pks value and the actual concentration of your solution. Hint: what is the concentration of 2 g of your unknown acid in 50 mL of water? Is this a solution of pure weak acid pure weak base, or a mixture of the two? ons for Member B: SFSU CHEM 216 92 Ruilibria

Explanation / Answer

I can answer the second part of the question easily. We need the observed pH value which you haven’t supplied. I will show you how.

1st. part:

Let the weak monoprotic acid be denoted as HA.

Weight of HA taken = 1.994 gm

Molar mass of acid = 136.58 gm/mol

Moles of acid taken = (1.994 gm)/(136.58 gm/mol) = 0.0146 mole

0.0146 mole of weak acid is dissolved in 50.0 mL distilled water. The molarity of the acid solution is [(0.0146 mol)/(50.0 mL)]*(1000 Ml/1 L)] = 0.292 M.

Therefore, we are titrating 0.292 M weak acid with 1 M NaOH solution. The reaction can be written as

HA (aq) + NaOH (aq) -----------> NaA- (aq) + H2O (aq)

Now, when 25% of the acid is neutralized, the moles of acid that undergoes reaction = (25/100)*(0.0146 mole) = 3.65*10-3 mole.

Moles of acid remaining = (0.0146 – 3.65*10-3) mole = 0.01095 mole

Moles of base formed = moles of acid reacted = 3.65*10-3 mole.

Now, we need to calculate the volume of NaOH used. Since NaOH and the weak acid react in a 1:1 molar ratio, hence,

moles of NaOH reacted = 3.65*10-3

Volume of NaOH required = (3.65*10-3 mole)/(1 mole/L) = 3.65*10-3 L = (3.65*10-3 L)*(1000 mL/1 L) = 3.65 mL.

Total volume of the solution = (50.00 + 3.65) mL = 53.65 mL

Concentration of base = [A-] = [(3.65*10-3 mole)/(53.65 mL)]*[(1000 mL/1 L)] = 0.0680 M

Concentration of unreacted acid = [HA] = [(0.01095 mole)/(53.65 mL)]*[(1000 mL/1 L)] = 0.2041 M

Ratio of [A-]/[HA] = (0.0680 M)/(0.2041 M) = 0.333

and log10[A-]/[HA] = log10(0.333) = -0.4775

Now, as per Henderson-Hasselbach equation,

pH = pKa + log10[A-]/[HA]

We need to know the pH to calculate the pKa. Please supply that information.

2nd. part:

The dissociation of the weak acid in water follows the equation

HA (aq) <======> H+ (aq) + A- (aq)

The initial concentration of HA is 0.292 M, as shown above. To calculate the initial pH, we need to set up the ICE chart as below:

HA (aq) <=======> H+ (aq) + A- (aq)

initial                             0.292                              0              0

change                            - x                               + x           + x

equilibrium             (0.292 – x)                           x              x

The equilibrium constant (acid dissociation constant, Ka) is given as

Ka = [H+][A-]/[HA] = (x)(x)/(0.292 – x)

Given pKa (literature value) = 7.2; therefore, Ka = 10-pKa [since pKa = -log10Ka]

= 6.3096*10-8

Therefore,

6.3096*10-8 = x2/(0.292 – x)

We need to make an assumption here; since Ka is small, we can assume (0.292 – x) 0.292. Thus,

6.3096*10-8 = x2/0.292

====> x2 = 1.8424*10-8

====> x = 1.3573*10-4

Therefore, [H+] = 1.3573*10-4 M and initial pH = -log10[H+] = -log10(1.3573*10-4) = -(-3.8673) = 3.8673 3.87 (ans)

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