Question

In a titration of 23.91 mL of 0.1347MH2​SO4​,42.41 mL of a NaOH solution are needed. What is the concentration (in inol/L) of the NaOH solution? Include only the numerical answer (no units).

Answer 1

0.1519

no units and to 4 sig figs

[M NaOH]

Step-by-step explanation:

We need to start with a balanced reaction:

H2SO4 + 2NaOH = Na2SO4 + 2H2O

The balanced equation tells us that we need 2 moles of NaOH for every one mole of H2SO4. We are starting with H2SO4 and adding NaOH until the reaction is completed.

We need to find the number of moles contained in 23.91 ml of 0.1347M H2SO4. Remember that M stands for (moles/liter). Lets use the original unit here. Also remember that to find moles, multiply the concentration, M, times the volume, V (making sure the units correctly cancel).

moles = M*V

moles H2SO4 = (0.1347moles/liter H2SO4)*(23.91 mL H2SO4)

Note that we need a conversion factor so that the volume units cancel properly:

moles H2SO4 = (0.1347moles/liter H2SO4)*(23.91 mL H2SO4)*(1 liter/1000ml)

moles H2SO4 = (0.1347moles H2SO4)*(23.91 H2SO4)*(1/1000) [cancel the units]

We are left with just moles, a good sign we're doing the calculation correctly.

H2SO4 = 0.003221 moles

From the balanced equation, we know we need twice the moles of NaOH, so we can write:

moles NaOH = 2*(0.003221 moles H2SO4)

moles NaOH = 0.006441 moles

Now we need to find the NaOH concentration. We have 0.006441 moles of NaOH in 42.41 ml. Write that as a concentration (moles/volume):

(0.006441 moles of NaOH)/( 42.41 ml) = 0.0001519 moles/ml NaOH

Convert this to the standard concentration unit of Molar (moles/liter) by using a conversion factor:

(0.0001519 moles/ml NaOH)*(1000ml/liter) = 0.1519 M NaOH