Problem Three A. 33) In a titration to determine the Alkalinity using 200 ml sam

Question

Problem Three A. 33) In a titration to determine the Alkalinity using 200 ml sample water and titrating with 0.02 N H2S04 the following species were identified. OH 7 mg/L, as calcium carbonate HCOs 100 mg/L, as calcium carbonate COs 50 mg/L, as calcium carbonate Answer the following: How many milliliters of acid were used in the first part of the titration to reach the first end point of 8.3? How many additional milliliters of acid was required to reach the final end point of pH 4.5? What was the initial pH of the sample? 1. 2. 3.

Explanation / Answer

First consider that the initial part of the titration ends up at the moment the hydroxide is neutralized. So, we can use the formula: C1V1 = C2V2, where C stands for concentration while V stands for volume.

Now in order to use this formula, you should first have the same units to work with. We can change the units of 0.02N, from the acid.

Remember that the formula for normality is: N= m (# of particles) / (Mol. Weight)(V)
0.02N = m (2particles) / V (98.078g/mol);
m/V = 0.02N(98.078g/mol)/ 2 particles = 0.98078 g/L = 980.79 mg/L

Now use that concentration to calculate the volume of acid needed:
(980.79mg/L)(V) = (7mg/L)(200ml)
Now so that this formula works, remember that you would actually need only half of the acid volume to neutralize the basic component in the solution. This is because you are titrating OH- molecules with an acid that has two hydrogens to neutralize them with. So you must calculate the volume of acid and then divide it over 2 to get the actual volume.

V = 1.427 ml / 2 = 0.71ml for the first part of the titration.

The following part is calculated similarly, just considering that you would also need half the volume for the HCO3 but the actual calculated volume for the CO3 because of how many hydrogens they need to neutralize.
** Remember posting each question separately, so they can all be answered :) **

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