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If 3 L of 1 M NaOH solution is prepared by mixing properties of stock solutions of $x$ L of 2.5 M NaOH and $y$ L of 0.4 M NaOH and no water is to be used, find $x : y$.

My attempt:

$$3 = 2.5x + 0.4y$$

Now I am stuck how to calculate $x:y$.

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In addition to the equation for the preservation of the amount of substance you've written, there is one more equation you can use for the system, namely the one reflecting preservation of volumes: $x + y = 3$:

$$ \left\{ \begin{align} x + y &= 3 \\ 2.5x + 0.4y &= 3 \end{align} \right. $$

$$x + y = 2.5x + 0.4y$$

$$1.5x = 0.6y$$

$$\frac{x}{y} = \frac{6\cdot 2}{10\cdot 3} = \frac{2}{5}$$

E.g. the answer is $x : y = 1 : 2.5$.

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As a quicker way, you could also use the Pearson's square to solve this:

1.0 M - 0.4 M = 0.6 M

2.5 M - 1.0 M = 1.5 M

The ratio is 0.6 to 1.5 (which corresponds to 1 to 2.5). Notice that we did not use the final volume in this calculation, the answer does not depend on it.

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