# A chemist titrates 300.0 mL of HCl with a 3.5 M solution of NaOH and requires 48.5 mL of the base to reach the endpoint. What is the concentration of the hydrochloric acid?

*print*Print*list*Cite

### 3 Answers

This is an example of a classic strong acid and strong base titration. Hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH) to form table salt (sodium chloride; NaCl) and water (H2O). The chemical equation is shown below:

HCl + NaOH --> NaCl + H2O

We can see that one mole of acid reacts with one mole of base. So if we can find the number of moles of NaOH used to reach the endpoint, then we know that this will be the same as the number of moles of HCl present to begin with. Since we know both the concentration and volume of NaOH used, we can readily calculate the moles by multiplying the two values.

(3.5 moles/liter) * (0.0485 liters) = 0.170 moles NaOH

Since the acid and base react in a 1 to 1 molar fashion, we know that this means that 0.170 moles of HCl was present in the original solution of hydrochloric acid. Since we know that 300 mL of acid was present, we can divide the two values to find the concentration.

(0.170 moles/0.3 liters) = 0.567 M HCl

**So the concentration of hydrochloric acid used is 0.567 M.**

**Sources:**

When solving for titration problems, we must first find out the mole ratios in which the acid and base react, in addition to the type of acid and base that is present (weak vs. strong).

In this case, the mole ratios are very simple. The reaction takes place as follows:

`HCl + NaOH -> H_2O + NaCl`

This shows that the two compounds react in a 1:1 ratio. Furthermore, they are both strong acids (HCl) and bases (NaOH), meaning they dissociate completely, so we do not have to worry about computing how much actually goes toward the reaction.

When asked to determine the molarity of an unknown solution given the volume used and the molarity and volume of the other solution, the following equation can be used:

`M_1V_1 = M_2V_2`

in which M = molarity and V = volume.

After making sure both the units for volume are the same, the numbers can simply be plugged into this equation.

`(3.5M)(300.0ml) = M_2(48.5ml)`

Then, solve for `M_2`

`M_2 = ((3.5M)(300.0ml))/(48.5ml) = 0.57M`

Thus, the molarity of the unknown solution is 0.57M

HCl and NaOH are strong acids/bases, which means they dissociate completely. Their ions are the same molarity as the solutions.

At the equivalence point, the moles of H+ equals the moles of OH-. You can use this equation:

`["acid"]("volume") = ["base"]("volume")`

Plug in your known values...

`["acid"](300"mL") = (3.5 "M")(48.5 "mL")`

... and solve for the unknown.

`["acid"] = 0.566 "M"`

Or 0.57M with two sig figs. There is no need to convert mL to L if the units are the same on both sides of the equal sign.