Unknown 0.05g of calcium into a 10mL of water and then add few drops of 1M HCl t

Question

Unknown 0.05g of calcium into a 10mL of water and then add few drops of 1M HCl to dissolves the Ca2+. The solution is transferred into a 250mL volumetric flask, add 2 drops of an indicator and few drops of NaOH until the solution turns pink. Lastly, add distilled water to the 250mL mark.

The calibration curve of the standards is y=20x+92 and the unknown has a potential of 60mV. Therefore the [ unknown Ca2+] = 0.02512M

y= mV, x = log[Ca2+]

The question did not say to measure the amount of HCl or NaOH added??

Determine the weigh percentage of Ca2+.

Determine the

Explanation / Answer

The calcium is generally seen in the form of calcium carbonate. it will react with water but will not completly dissolved in water. To make it dissolve in water HCl is added. When HCl is added, Calcium carbonate reacts with it and forms calciumchloride. It is This calcium Chloride which is Etsimated in this procedure. We also add NaOH during the procedure This is to nuetrilize the unreacted HCl in the solution.

A calibration curve of y = mx +c is prepared using the standard. The sample is analysed and obtained a potetial of 60 mV

Here according to the procedure of ion selective electrode E = constant + slope * log [Concentration of ca2+] which is in the form of y = mx +C ( where E = potential measured)

60 = 20 x + 92

60-92 = 20 x

-32 /20 = x

x = -1.6

1.6 = log x

x = antilog 1.6

= 0.02512 M

The question is about the estimation of calcium in the solution, because finally there is no HCl or NaOH in the solution, Because HCl added reacted with caCO3 and formed Calcium Chloride and NaOH added nuetrilized the excess HCl and formed water and salt. The potential produced is only due to Calcium . That is why question did not ask about HCl or NaOH added initially.

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