Obtaining 30% sulfuric acid [closed]

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?

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.

closed as off-topic by Mithoron, airhuff, Tyberius, Jon Custer, Todd MinehardtDec 8 '17 at 0:53

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• Hello and welcome to Chemistry.SE. This is a homework type of question. We have a policy which states that ‎you should show your thoughts and/or efforts into solving the problem. It'll make us certain that ‎we aren't doing your homework for you. Otherwise, this question may get closed.‎ Please edit in your full reasoning or thoughts on this. Best of luck... – airhuff Dec 7 '17 at 21:08
• Hello, I don't need the answer or the solution using the same exact units, the problem lies in me not finding an example problem which has the 2nd solution having units and the finished one not having units (by units I mean volume, mass etc.) – Kurai Dec 7 '17 at 21:14
• Let's discuss basics. What is a 10% solution of sulfuric acid? – MaxW Dec 7 '17 at 21:26
• 10% solution means that it has a procentile mass of 10% sulfuric acid mixed with water, the same being with 50%, so by finding the mass of the whole solution 300 and multiplying it by ρ of the selected solution you get the whole mass of the solution which is 420 and that multiplying by 0.5(50%) you get 210grams of H2SO4 – Kurai Dec 7 '17 at 21:33

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:

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 $V_1$:

$$V_1 = \frac{\rho_2 (\omega_2 - \omega_3)}{\rho_1 (\omega_3 - \omega_1)} \cdot V_2 = \frac{\pu{1.40 g cm-3} \cdot (0.5 - 0.3)}{\pu{1.07 g cm-3} \cdot (0.3 - 0.1)} \cdot \pu{300 mL} = \pu{392.5 mL}$$

Molar and equivalent concentrations

In the comments @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?

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:

$$V_1 = \frac{C_2 - C_3}{C_3 - C_1} \cdot V_2 = \frac{C_\mathrm{eq2} - C_\mathrm{eq3}}{C_\mathrm{eq3} - C_\mathrm{eq1}} \cdot V_2 = \frac{\pu{10 N} - \pu{6 N}}{\pu{6 N} - \pu{2 N}} \cdot \pu{300 mL} = \pu{300 mL}$$

• Methinks this might goes slightly over the OP's head. But it is correct of course. Nicely formatted too. – Gert Dec 7 '17 at 22:07
• @Gert Thank you, I've thought of graphical method, but in the end there is still some math to handle, so I decided to go with this one. Do you think I should've included intermediate steps when expressing $V_1$ for clarity? – andselisk Dec 7 '17 at 22:10
• (Even though it says to avoid using comments to thank people, I just had to do so(don't have 15 rep. points)) Thank you, a lot. My problem was that the formula I knew couldn't be used like that, so basically I had to figure out the solution by guessing. Is there a place where I could learn to extrapolate formulas out of other formulas that big? Thank you, once again for editing my post and also answering my question. – Kurai Dec 7 '17 at 22:10
• @Kurai You are very welcome:) I'm not quite sure I understood what you mean by "extrapolate", you basically take the basic formula for the corresponding concentration (mass % in this case) and try to express the unknown parameters using the equations for basic physical quantities, such as density or molecular weight. And yes, when given "$x\%$ solution of compound" without further notes, it's usually safe to assume it's mass fraction $\omega$. – andselisk Dec 7 '17 at 22:17
• What if the question had been formulated in the following way. How much 2N H2SO4 must be added to 300mL of 10N H2SO4 to prepare a 6N H2SO4 solution? – user55119 Dec 7 '17 at 23:42