# Titration calculations - working out cation [closed]

so I was doing this question and I have some questions about what I did. Firstly am I correct to say that the concordant results = 24.1 cm^3 because there are 2 of these values. From this I then calculate that M = 32.9 however this can not be true, so what have I done wrong?  • Let us start by writing the balanced equations. Write a balanced equation showing the reaction of NaOH and HCl. – M. Farooq Oct 13 '20 at 2:59
• Be aware the intention of Chemistry SE site is not task solving nor proof reading service, but rather help in understanding underlaying principles. Is there any principle or idea you have trouble to apply ? Also,write in detail your calculations. How can we tell you what you did eventually wrong if you do not show us what you did ? – Poutnik Oct 13 '20 at 3:50

You must have forgotten a factor $$2$$ somewhere. You are right is thinking that $$24.10$$ mL is needed for the final titration. But now let's redo the calculations.

$$24.10$$ mL $$\ce{NaOH} ~0.25$$ M contains $$24.10·0.25 = 6.025$$ millimole $$\ce{NaOH}$$. Obviously this $$\ce{NaOH}$$ has neutralized the same amount $$6.025$$ millimole $$\ce{HCl}$$. As a consequence, there was $$10$$ times more $$\ce{HCl}$$ in the $$250$$ mL flask. This is $$0.06025$$ mol $$\ce{HCl}$$, and was the residual amount of $$\ce{HCl}$$ remaining after reaction of HCl plus the unknown carbonate. It also means that after adding $$0.100$$ mole $$\ce{HCl}$$ in the initial flask, the difference $$0.100 - 0.06025 = 0.03975$$ mol $$\ce{HCl}$$ had been used to destroy the carbonate.

But whatever the formula of the initial carbonate, $$2~\ce{H+}$$ are needed to react with $$1~ \ce{CO3^{2-}}$$. So the amount of unknown $$\ce{XCO3}$$ or $$\ce{Y2CO3}$$ carbonate in the original powder is half the amount of $$\ce{HCl}$$ and it is : $$0.03975/2$$ mol = $$0.019875$$ mol $$\ce{XCO3}$$.

As the mass of this carbonate is $$2.50$$ g, its molar mass is: M = m/n = $$2.50/0.019875 = 125.8$$ g/mol. As the molar mass of $$\ce{CO3}$$ is $$60$$ g/mol, it means that the molar mass of $$\ce{X}$$ is $$125.8 - 60 = 65.8$$ g/mol.

This final value may be twice the atomic mass of a monovalent cation, or the atomic mass of a bivalent cation. As there are no monovalent cation with an atomic mass of about $$32.9$$, the unknown element must be a bivalent cation. And the only element with such an atomic mass is Zinc : A(Zn) = $$65.39$$

So the unknown carbonate is $$\ce{ZnCO3}$$

As an analytical chemist I disagree with Maurice's statement that 24.10 is the "right" value for the titrant. Solving for (d) does given a value of 24.10 ml yielding (1) 24.10, (2) 24.45 and (3) 24.10. Getting two values of 24.10 must be viewed as a coincidence. Thus the average of 24.22 should be used. (I'd use 24.2167 in the intermediate calculations to reduce rounding errors.)

So you need to calculate the standard deviation from the three measurements and then use the Student's T test to calculate a 95% confidence interval about the mean. Finally calculating the 95% confidence interval as a % error value will give you the expected error around the final mass.

For the three values the mean is 24.22 and the standard deviation is 0.20. For a two-tailed Student's T test at 95% CI and two degrees of freedom, the T value is 4.3027. Thus the CI is +/- $$(\frac{0.2}{\sqrt{3}}\times4.3027 = 0.50)$$. Looking at this as a percentage gives +/- 2.06% for the calculated cation mass value.

Finally the problem statement that the sample is an "unknown carbonate" is somewhat ill-defined.

• The problem doesn't specifically state that there is only a single cation. It could for example be the mineral dolomite, $$\ce{CaMg(CO3)2}$$.
• Nor does the problem state that the cation has a single oxidation state. There are minerals with Fe(II)/Fe(III).
• The problem also doesn't state that the salt isn't a hydrate.

The point is that you must make an assumption that the mineral is not a hydrate and either a monovalent cation, $$\ce{X2CO3}$$, a divalent cation, $$\ce{YCO3}$$ or perhaps a trivalent cation $$\ce{Z2(CO3)3}$$.

• This original question is taken from a British chemistry preparation book for pre-university students. I don't think the question writer ever thought to make it very complicated to have mixed valence or mixed carbonates. – M. Farooq Oct 13 '20 at 20:10
• @ MaxW. The difference between my value (24.10 mL), and yours (24. 22 mL) is about 0.5%. This is also the precision of the original data : 2.50 g is supposed to be 2.50 ± 0.01 g, which is also ± 0.5%. On the other hand, I don't know of any carbonate containing a triply charged metallic ion. These ions are too acidic in solution. Al, Fe(III) and Cr(III) do not form carbonates. – Maurice Oct 13 '20 at 20:57

I agree with MaxW about poorly worded question. This should at least states that the unknown carbonate is anhydrous. Assuming it is an anhydrous carbonate with $$\ce{M2(CO3)_n}$$ formula, I did the calculations and found out that: $$M=32.9n$$ where $$M$$ is the atomic mass of the metal cation with $$n+$$ charge ($$\ce{M^n+}$$). Thus, OP's result of $$M=32.9$$ probably obtained by assuming the unknown carbonate is an alkali carbonate with molecular formula of $$\ce{M2CO3}$$ where $$n=1$$. However there is no metal with $$+1$$ charge with $$32.9$$ atomic mass (closest is $$\ce{K+}$$, atomic mass which is $$39.1$$). The next best choice is when $$n=+2$$, thus $$M=65.8$$. The closest $$\ce{M^2+}$$ is $$\ce{Zn^2+}$$, atomic mass of which is $$65.4$$. This is an acceptable value. Still, if you want to consider the possibility of $$n$$ being equals to $$+3$$ (thus $$M=98.7$$), the closest $$\ce{M^3+}$$ is $$\ce{Tc^3+}$$, atomic mass of which is $$98.9$$. Although this is slightly better value than $$\ce{Zn^2+}$$, I doubt whether $$\ce{Tc2(CO3)3}$$ is exist or not (although $$\ce{TcCl3}$$ is exist). Therefore it is safe to say that the unknown carbonate we are dealing with here is more like $$\ce{ZnCO3}$$.

• I also that Zn is the most "suitable" answer. The posted image is from a test-prep book for pre-university students. They are neither analytical chemists nor they are aware of Tc compounds or mixed valence compounds. – M. Farooq Oct 13 '20 at 20:13