I have this question:
A 71.7-mL solution of 0.182 M NaOH is titrated with 0.2086 M HC.
And I got the rest of the question of right...but for this one part.
4.) after addition of 50.0 mL of HCl
i keep on getting 12.95 but I'm wrong...can someone tell me what to do?
Volumebase*molaritybase=volumeacid*molarityacid
Volume acid= 71.7ml*.182M/.2086M
...and what do I do withe 50?
*with the
To determine the concentration of NaOH in the solution after the addition of HCl, we can use the concept of stoichiometry. Stoichiometry is the calculation of quantities of reactants and products in a chemical reaction. In this case, the stoichiometry is given by the balanced chemical equation:
NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l)
From the equation, we can see that the ratio between NaOH and HCl is 1:1. This means that 1 mole of NaOH will react with 1 mole of HCl.
To find the moles of HCl added, we can use the formula:
moles = concentration (M) × volume (L)
In this case, the concentration of HCl is 0.2086 M, and the volume is 50.0 mL (which needs to be converted to liters by dividing by 1000):
moles of HCl = 0.2086 M × (50.0 mL / 1000) L = 0.01043 moles
Since the stoichiometry is 1:1, the moles of NaOH present in the solution will also be 0.01043 moles.
Now, to calculate the final concentration of NaOH, we need to consider the final volume of the solution. Initially, we had a 71.7 mL solution of NaOH. After adding 50.0 mL of HCl, the total volume will be:
71.7 mL + 50.0 mL = 121.7 mL
To convert this to liters, we divide by 1000:
volume (L) = 121.7 mL / 1000 = 0.1217 L
Finally, we can use the formula to find the final concentration:
concentration (M) = moles / volume
concentration (M) = 0.01043 moles / 0.1217 L = 0.0855 M
Therefore, the final concentration of NaOH in the solution after the addition of 50.0 mL of HCl is 0.0855 M.
It seems like you might have made an error in your calculations. Remember to convert all units to the appropriate form (e.g., L) and ensure that your arithmetic is accurate.