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sp2606 [1]
3 years ago
10

Why is acid base titration a useful technique in chemistry

Chemistry
1 answer:
Elina [12.6K]3 years ago
7 0

Answer: Titration is particularly useful if we want to find out the amount or concentration of a known acid or base in a given sample.

Explanation: Suppose you have just been handed a beaker containing a fixed volume of HCL solution. If you simply just know the volume of the solution, you can titrate it against a reagent (base) to figure out the exact concentration of the solution (also known as molarity)

The solution is placed in a flask for titration and a minute amount of indicator is then added into the flask. The reagent (base) is placed in a burette and slowly added to the solution and indicator mixture. The amount of reagent used is recorded when the indicator causes a change in the color of the solution. Using Stoichiometry, we are then able to calculate the concentration and moles of the HCL solution used up in the reaction.

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Using the brønsted theory, classify the compounds as either an acid or a base
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8 0
3 years ago
If you have 100 ml of a 0.10 m tris buffer (pka 8.3) at ph 8.3 and you add 3.0 ml of 1.0 m hcl, what will be the new ph?
sukhopar [10]

The new pH is 7.69.

According to Hendersen Hasselbach equation;

The Henderson Hasselbalch equation is an approximate equation that shows the relationship between the pH or pOH of a solution and the pKa or pKb and the ratio of the concentrations of the dissociated chemical species. To calculate the pH of the buffer solution made by mixing salt and weak acid/base. It is used to calculate the pKa value. Prepare buffer solution of needed pH.

                       pH = pKa + log10 ([A–]/[HA])

Here, 100 mL  of  0.10 m TRIS buffer pH 8.3

                 pka = 8.3

         0.005 mol of TRIS.

∴  8.3 = 8.3 + log \frac{[0.005]}{[0.005]}

<em>    </em>inverse log 0 = \frac{[B]}{[A]}

   \frac{[B]}{[A]} = 1

Given; 3.0 ml of 1.0 m hcl.

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           0.003 mol of HCL.

pH = 8.3 + log \frac{[0.005-0.003]}{[0.005+0.003]}\\pH = 8.3 + log \frac{[0.002]}{[0.008]}\\\\pH = 8.3 + log {0.25}\\\\pH = 8.3 + (-0.62)\\pH = 7.69

Therefore, the new pH is 7.69.

Learn more about pH here:

brainly.com/question/24595796

#SPJ1

 

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