# Battery acid solution preparation from 98% ACS Sulfuric Acid

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.

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.

• Comments are not for extended discussion; this conversation has been moved to chat. Jul 4, 2020 at 9:03

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.