# How to identify the ratio of oxidation states in terbium ions in its mixed oxide?

Estimate the ratio of $$\ce{Tb^3+}$$ to $$\ce{Tb^4+}$$ in $$\ce{Tb_6O_11}$$.

For my solution I knew that I had to get a sum of 22 positive charges from 6 $$\ce{Tb^3+}$$ and $$\ce{Tb^4+}$$ ions, so I just added different combinations until I found that $$4(\ce{Tb^4+})+2(\ce{Tb^3+})$$ gave me the desired value to have an overall neutral oxidation state on the molecule. This gives a $$2:1$$ ratio also. Is this the correct way to do this, or is there a more formal method?

• your way looks good to me Dec 15 '15 at 13:20
• Well, trying different combinations is definitely a lot better than just sit there and stare at it. You could go with a linear equation, but with numbers like these it is hardly any faster. Dec 15 '15 at 13:20

There is certainly a formal way. Let $x$ and $y$ be the number of $\ce{Tb^3+}$ and $\ce{Tb^4+}$ ions respectively. Then:
\begin{align} 3x + 4y &= 22 \\ x + y &= 6 \end{align}
\begin{align} \begin{pmatrix} 3 & 4 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} 22 \\ 6 \end{pmatrix} \\ \begin{pmatrix} x \\ y \end{pmatrix} &= \frac{1}{3\cdot 1 - 1\cdot 4} \begin{pmatrix} 1 & -4 \\ -1 & 3 \end{pmatrix} \begin{pmatrix} 22 \\ 6 \end{pmatrix} \\ &= - \begin{pmatrix} 1\cdot 22 - 4\cdot 6 \\ -1 \cdot 22 + 3\cdot 6 \end{pmatrix} \\ &= \begin{pmatrix} 2 \\ 4 \end{pmatrix} \end{align}