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I'm trying to prepare some battery acid for activating a flooded lead acid battery I had purchased. The battery concentration should be around 36-28% sulfuric acid solution. I have decided to go with 37% acid solution. I would like to confirm if the volume of acid to be added is correct. So, using a 98% ACS reagent sulfuric acid the volume of acid to make 100mL solution should be 37.57% right? The remaining volume is distilled/deionized water to make up for the 100mL.

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2 Answers 2

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The answer begins with the general calculation of mixing solutions with various densities. If you are interested only in the particular numeric results, or if you are scared by algebra, , skip the theoretical part between the horizontal lines.


There is the mixing cross rule, deriving the ratio of mass of 2 mixed liquids from their mass percentage.

$$\frac { m_2 }{m_1} = \frac{|w_1 - w_0|}{|w_2 - w_0|}=k\tag{1}$$ respectively $$m_2 = k\cdot m_1 \tag{2}$$ and $$m_0 = m_1 + m_2 = m_1 + k \cdot m_1 = m_1 \cdot \left(1 + k \right) \tag{3}$$

where
$m_1,m_2, m_0$ are respective masses of mixed solution 1 and 2 and ofthe final solution 0
$w_1,w_2,w_0$ are respective mass percentages of mixed solutions 1 and 2 and the final solution 0

As we are interested rather in volumes, we involve solution densities:
$$V_1 = \frac{m_1 }{\rho_1}\tag{4}$$ $$V_2 = \frac{m_2 }{\rho_2} = \frac{m_1 }{\rho_2} \cdot k\tag{5}$$ $$V_0 = \frac{m_0 }{\rho_0} = \frac{m_1 \cdot \left(1 + k \right) }{\rho_0}\tag{6}$$

We have the given final volume $V_0$ of the final solution of the density $\rho_0$.

Then, for the solution 1:

$$m_1 = V_0 \cdot \rho_0 \cdot \frac{1}{\left(1 + k \right)} \tag{7}$$ respectively the volume: $$V_1 = V_0 \cdot \frac{ \rho_0}{\rho_1 } \cdot \frac{1}{1 + k}\tag{8}$$

For the solution 2:

$$m_2 = m_1 \cdot k = V_0 \cdot \rho_0 \cdot \frac {k}{1+k} \tag{9} $$

respectively the volume:

$$V_2 = V_0 \cdot \frac{\rho_0}{\rho_2 } \cdot \frac {k}{1+k} \tag{10} $$


Applying the particular data:

Solution 0: 37% $\ce{H2SO4}$,The density $\pu{1.267 g/mL}$, target volume $V_0 = \pu{100 mL}$

Solution 1: 98% $\ce{H2SO4}$.The density $\pu{1.8361 g/mL}$

Solution 2: water = 0% $\ce{H2SO4}$, the density $\pu{1.000 g/mL}$

From the equation (1): $k = \frac{|98-37|}{|0-37|}=61/37 \simeq 1.649$

We plug in the data into equations (8) and (10)

For 98% $\ce{H2SO4}$: $$V_1 = \pu{100 mL} \cdot \frac{ \pu{1.267 g/mL}}{\pu{1.8361 g/mL} } \cdot \frac{1}{1 + 1.649} \simeq \pu{26 mL}\tag{8a}$$

For water: $$V_2 = \pu{100 mL} \cdot \frac{ \pu{1.267 g/mL}}{ \pu{1.000 g/mL}} \cdot \frac {1.649}{1+1.649} \simeq \pu{79 mL} \tag{10a} $$

Be aware volumes are not additive. After cooling down, the total initial volume shrinks from $\pu{105 mL}$ to $\pu{100 mL}$.

For subsequent dilutions, it is an advantage to keep some volume of the residual 37% solution and to do the dissolution in 2 steps.
First, dilute the calculated portion of 98% H2SO4 in this rest of 37% H2SO4. After cooling, pour this mixture to the calculated volume of water. The amount of heat per dilution would be divided into these 2 steps.

DO NOT forget to pour acid into water, as not vice versa !! Pouring water into acid would very propably end by a hot acid splash into your face, as mixing is very exothermic. Mix them carefully, slowly, continually mixing and cool it by a water bath. Pour the acid along the glass rod and not directly. Wear a lab coat or at least an apron, gloves and shield if possible. Be aware even tiny drops can cause after some days/weeks holes in your clothes.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – andselisk
    Jul 4, 2020 at 9:03
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I upvoted the answer by @Poutnik, but here is a slightly different take. I assume the 98% and 37% concentrations are by mass, i.e., 100 g of 98% concentrated sulfuric acid contains 98 g of pure acid plus 2 g of water. I also assume the 98% and 37% are exact, to avoid initial fussing with significant figures. Round off to two digits is near the end.

Take 100 g of the 98% concentrated sulfuric acid. It contains 98 g of pure acid, which is to be 37% of the diluted acid. Since 98 is 37% of 264.865, the mass of water required is 164.865 g, i.e., 264.865 g total minus 100 g of 98% sulfuric acid.

From Poutnik's answer, the density of the 37% acid is 1.267 g/mL. Hence the total volume of the 37% acid is 209.05 mL. Since the OP wanted 100 mL of 37% acid, just scale down by 100/209.05, which is 0.4784. Therefore 47.84 g of 98% sulfuric acid is slowly added to 78.864 g of water. For the water, 79 mL is an adequate approximation. For the 98% sulfuric acid, the density is 1.8361 g/mL, as per Poutnik's answer. Hence the volume required is 47.84 g divided by 1.8361 g/mL, which is 26.055 mL. So 26 mL is an adequate approximation.

Summary: As per the safety precautions given in Poutnik's answer, 26 mL of 98% concentrated sulfuric acid is slowly added (with stirring) to 79 mL of water.

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