2. Repeat Part D using 25.00 mL of the unknown weak acid (instead of HCl). The u

Question

2. Repeat Part D using 25.00 mL of the unknown weak acid (instead of HCl). The unknown l have a concentration of approximately 0.1 M. The equivalence point of a weak acid wil acid/strong base titration is at an alkaline pH.Qureumin, which changes color within the range 8-9, can be used as the indicator. 3. Plot pH vs. volume of NaOH added. Find the concentration of the unknown acid and values for pKa and Ka. Recall that the pKa corresponds to the pH at half the equivalence point yolume. Be sure to gather some pH data around this volume. lunc A 3.3 3, 7 4.2 1.3 2 5. 2 5.3 s.f 2.9 20 24( 2f5 26/

Explanation / Answer

First of all we have to find the concentration of unknown acid.

Given,

Initial concentration of unknown acid = 0.1 M

Volume of unknown acid used = 25 ml

Reaction of NaOH with unknown acid,

HA + NaOH ---> NaA + H2O

1 mol of unknown acid reacts with NaOH to form 1 mol conjugate base of the acid (A-). So, by adding NaOH to the solution decreases the moles of unknown acid (concentration).

To find the concentration of the unknown acid,

Initial No. of moles of unknown acid = Molarity x Volume

= 0.1x 25= 25 mmol

When 1 ml of NaOH is added. So,

No. of mole of NaOH added = 0.1008 x 1= 0.1008 mmol

{Molarity of NaOH = 0.1008}

Now,

No. of moles of unknown acid left = Initial No. of moles of unknown acid - No. of moles of NaOH

= 2.5 - 0.1008= 2.3992 mmol

Total volume of solution = 25 ml + 1ml = 26 ml

So, concentration of unknown acid = no. of mol/volume of solution

= 2.3992 mmol/ 26 ml = 0.0923 mmol/ml= 0.0923 M

Similarily doing for other values, we get the concentration of unknown acid in each steps.

8.7

We know that

pH = pKa + log([A-]/[HA])

when [A-]=[HA]

then pH=pKa

As the volume of the solution is same. So we can deal with no. of moles of [A-] and [HA].

and we know

No. of moles of NaOH consumed = No. of moles of A- produced.

So, from the table, we clearly see that

at pH =4.8

the no. of mole of NaOH is approximately equal to no. of moles of unknown acid.

So,

pKa = 4.8

pKa = -log Ka

Ka = 10^(-4.8) = 1.58 x 10^(-5)

Volume of NaOH (ml) pH No.of moles of NaOH(mmol) No. of moles of unknown acid (mmol) Concentration of unknown (M) 0 3.3 0 2.5 0.1 1 3.7 0.1008 2.3992 0.0923 2 3.9 0.2016 2.2984 0.0851 3 4.1 0.3024 2.1976 0.0785 4 4.2 0.4032 2.0968 0.0723 5 4.3 0.5040 1.996 0.0665 6 4.4 0.6048 1.8952 0.0611 7 4.5 0.7056 1.7944 0.0561 8 4.6 0.8064 1.6936 0.0513 9 4.6 0.9072 1.5928 0.0468 10 4.7 1.0080 1.492 0.0426 12 4.8 1.2096 1.2904 0.0349 14 4.9 1.4112 1.0888 0.0279 16 5.1 1.6128 0.8872 0.0216 18 5.2 1.8144 0.6856 0.0159 20 5.3 2.0160 0.484 0.0108 22 5.6 2.2176 0.2824 0.0060 24 5.8 2.4192 0.0808 0.0016 25 6.3 2.5200 -0.02 - 25.5 7.9 2.5704 -0.0704 - 26

8.7

2.6208 -0.1208 -

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.