Obtaining 30% sulfuric acid [closed]








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> 
> What volume of $10\%$ sulfuric acid solution ($\rho = \pu{1.07 g cm-3}$) do you need to put into $\pu{300 mL}$ $50\%$ of the same acid solution ($\rho = \pu{1.4 g cm-3}$) to get $30\%$ solution?
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I've been searching for a solution for the past few hours, I've seeken out the same kind of problem and found only one which had water in it, not the same solution. So if anyone was kind enough to show a solution I'd be enlightened.


I tried "Pearsons letter", which gives me that it has a ratio of $1 : 1$ for making 30% solutions for some reason (doesn't make sense, because then it'd be 300+300 when the answer is 392.5), I also tried making an algebraic formula, which gave me an answer that was about $60000$ units above the actual answer. I only need the solution, so I could solve the same kind of problems.


P. S. English isn't my first language, if you need me to specify something on the problem, just say. Thank you in advance.




First off, solving these kind of problems we are assuming that


1. the resulting mixture is homogeneous;
2. the mixing of various solutions takes place instantaneously;
3. the volume of the mixture is equal to the sum of the volumes of the miscible solutions.


The approach is very similar, there slight deviations based of how the concentrations are expressed. I suggest to start by writing down the equation for the given fraction, and then expand it using the provided physical quantities, finally expressing the volume. A couple of examples:




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### Mass fractions


This addresses the original question. By definition mass fraction $\omega\_i$ or $i$-component with mass $m\_i$ in solution is determined as follows:


$$\omega\_i = \frac{m\_i}{\sum\_i{m\_i}}$$


At the same time mass is expressed *via* density $\rho$ and volume $V$:


$$m = \rho V$$


Now, the resulting mass fraction $\omega\_3$ is:


$$\omega\_3 = \frac{\omega\_1 m\_1 + \omega\_2 m\_2}{m\_1 + m\_2} = \frac{\omega\_1 \rho\_1 V\_1 + \omega\_2 \rho\_2 V\_2}{\rho\_1 V\_1 + \rho\_2 V\_2}$$


All you have to do now is solve this equation for $V\_1$:


$$V\_1 = \frac{\rho\_2 (\omega\_2 - \omega\_3)}{\rho\_1 (\omega\_3 - \omega\_1)} \cdot V\_2 = \frac{\pu{1.40 g cm-3} \cdot (0.5 - 0.3)}{\pu{1.07 g cm-3} \cdot (0.3 - 0.1)} \cdot \pu{300 mL} = \pu{392.5 mL}$$




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### Molar and equivalent concentrations


In the comments [@user55119](https://chemistry.stackexchange.com/users/55119/user55119) suggested a similar problem and expressed a doubt about the necessity of using density in the OP's original question:



> 
> How much $\pu{2 N}$ $\ce{H2SO4}$ must be added to $\pu{300 mL}$ of $\pu{10 N}$ $\ce{H2SO4}$ to prepare a $\pu{6 N}$ $\ce{H2SO4}$ solution?
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By definition molar and equivalent concentrations are tied *via* the number of equivalents $z$:


$$C = \frac{C\_\mathrm{eq}}{z} = \frac{n}{V}$$


The resulting molar concentration is:


$$C\_3 = \frac{n\_1 + n\_2}{V\_1 + V\_2} = \frac{C\_1 V\_1 + C\_2 V\_2}{V\_1 + V\_2}$$


Solving the equation for $V\_1$ and taking into account that the equivalence factor $z^{-1}$ is a proportionality coefficient:


$$V\_1 = \frac{C\_2 - C\_3}{C\_3 - C\_1} \cdot V\_2 = \frac{C\_\mathrm{eq2} - C\_\mathrm{eq3}}{C\_\mathrm{eq3} - C\_\mathrm{eq1}} \cdot V\_2 = \frac{\pu{10 N} - \pu{6 N}}{\pu{6 N} - \pu{2 N}} \cdot \pu{300 mL} = \pu{300 mL}$$


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