# What is the necessary volume of concentrated HCl to prepare a solution with a pH of 1.8?

I am trying to solve the following exercise:

What volume of a concentrated $$\ce{HCl}$$ solution, which is $$36.0\%$$ $$\ce{HCl}$$ by mass and has a density of $$1.179~\mathrm{g~mL^{-1}}$$, should be used to make $$5.00~\mathrm{L}$$ of an $$\ce{HCl}$$ solution with a $$\mathrm{pH}$$ of $$1.8$$.

• I found the respective volumes that $$\ce{H2O}$$ and $$\ce{HCl}$$ contributed.
• I then found the molarity of the $$\ce{HCl}$$ aqueous solution given.
• I used $$10^{-\mathrm{pH}}$$ to get the amount concentration of $$\ce{HCl}$$ needed, I then proceeded to find the amount of substance of $$\ce{HCl}$$ by multiplying the concentration needed by $$5.00~\mathrm{L}$$. I then took the amount of substance of $$\ce{HCl}$$ needed and multiplied by the density given to get an answer of $$2.4~\mathrm{L}$$.

Am I correct?

Unfortunately no, you're not right. As mentioned in a now deleted comment by Bruno, your answer gives a $$\mathrm{pH}$$ of $$-0.74$$, which is not correct. You didn't give detailed steps of your calculations, and your first step doesn't seem like anything we should do.

We can solve this by making the calculations step by step. Let's start with the concentration of $$\ce{H+}$$, which will be the same as the concentration of $$\ce{HCl}$$, we will achieve:

$$\ce{[H+]} = 10^{-\mathrm{pH}} = \pu{0.01584893 mol//L}$$

Since we're going to prepare 5 liters, this means the amount of substance of $$\ce{H+}$$ should be

$$5 \times \pu{0.01584893 mol} = \pu{0.07924466 mol}$$

To transform this into amount of $$\ce{HCl}$$, we need the acid's molar mass ($$\pu{36.46 g//mol}$$) to convert from amount of substance to mass:

$$\require{cancel}% g_{\ce{HCl}} = \mathrm{\frac{36.46\ g}{\cancel{mol}}} \times 0.07924466\ \mathrm{\cancel{mol}} = 2.88926 \ \mathrm{g}$$

Now that we have the amount of substance of $$\ce{HCl}$$, we calculate the volume, which is where the density comes in:

$$V_{\mathrm{HCl}} = \mathrm{\frac{2.88926\ \cancel{g}}{1.179\ \cancel{g}\ mL^{-1}}} = 2.450602 \ \mathrm{mL}$$

Now the last step: Our stock solution isn't pure $$\ce{HCl}$$, it's diluted in water. So actually the volume we need is:

$$V_{\mathrm{HCl}} = 2.450602 \times \frac{100}{36} = \pu{6.807229 mL} \approx \pu{6.8 mL}$$

This might look that it is not much, but remember $$\ce{HCl}$$ is a very strong acid. For example, to prepare $$\pu{1000 mL}$$ of a $$\pu{0.001 M}$$ solution you only need $$\pu{0.086 mL}$$ of concentrated $$\ce{HCl}$$!

You can check this result by imputing the given density, weight percentage, final volume and $$\ce{H+}$$ concentration (first step) into Sigma-Aldrich's Molarity Calculator.

1.000 L of 36.0 % HCl with the density of 1.179 g/ml will weigh 1179 g and will contain 0.36*1179 g of HCl or 0.36*1179/36.46 = 11.641 mole of HCl. At pH = 1.8 => [H3O+] = 0.015849 (M). To make 5.0 L solution we need 5.0 * 0.015849 = 0.079245 (mole). Suppose we need x liter of the conc. HCl solution: x*11.641 = 0.079245 => x = 0.006807 (L) Answer: 6.8 ml.

It's simple. We all know that density = Mass/volume. The density of your hcl is 1.179

Mass of hcl in one molar solution which is the molar mass is 36.5

but the percentage purity of the hcl is 35% according to you. That means, 35% of that 36.5g is in 1 molar/dm3.

Which means, your mass will be 35% of 36.5 =36.5*35/100 35.5*0.35=12.775

Therefore, 1.179=12.775/v v=12.775/1.179 v=10.84 (approx)

This is the volume of your conc hcl needed to prepare 1M in 1 dm3

But you want to prepare 5 dm3 5 dm3=10.84*5=54.20

The simple formula that can guide you is

Molar mass times percentage purity everything over density

## MM x pp

D The volume u get using the above formula will prepare 1M in 1 dm3 so, you can now multiply by the number of molar concentration u want to prepare. Spasiba