If a 28 mLs of a 0.25 molar NaOH solution is used to titrate 14 mLs of a nitric acid to its phenolphthalein endpoint, what is the molar concentration of the acid?
mL x M = mL x M (Technically this is the dilution equation but you may use it in titrations in which the ratio of the titrant to the unknown solution is 1:1).
Well, it seems like you've stirred up quite the chemistry concoction there! Let's see if I can help you out while adding a little laughter to the equation.
To figure out the molar concentration of the nitric acid, we'll need to use the concept of stoichiometry – it's like the chemistry version of a balancing act!
First, we'll calculate the number of moles of NaOH using the formula:
moles of NaOH = volume of NaOH (in liters) * molar concentration of NaOH
Converting 28 mL of NaOH to liters, we get:
28 mL = 0.028 L
Now, to find the moles of NaOH, we multiply the volume by the molar concentration:
moles of NaOH = 0.028 L * 0.25 mol/L
Which gives us:
moles of NaOH = 0.007 mol
Since NaOH and HNO₃ react in a 1:1 ratio, the moles of HNO₃ will also be 0.007 mol.
Now, we need to calculate the molar concentration of the nitric acid. We do this by dividing the moles of HNO₃ by the volume in liters:
molar concentration of HNO₃ = moles of HNO₃ / volume of HNO₃ (in liters)
So, dividing 0.007 mol by 0.014 L (since 14 mL equals 0.014 L), we find:
molar concentration of HNO₃ = 0.007 mol / 0.014 L
Resulting in a molar concentration of:
molar concentration of HNO₃ = 0.5 mol/L
There you have it! The molar concentration of the nitric acid is 0.5 mol/L. Now, wasn't that a titillating way to find the answer? Keep those chemistry jokes coming!
To find the molar concentration of the nitric acid, we can use the stoichiometry of the reaction between NaOH and HNO3:
NaOH + HNO3 -> NaNO3 + H2O
The balanced chemical equation shows that one mole of NaOH reacts with one mole of HNO3.
First, let's calculate the number of moles of NaOH used in the titration:
moles of NaOH = volume of NaOH (in L) * molar concentration of NaOH
volume of NaOH = 28 mL = 28/1000 L = 0.028 L (convert mL to L)
molar concentration of NaOH = 0.25 mol/L
moles of NaOH = 0.028 L * 0.25 mol/L = 0.007 mol
Since the stoichiometry of the reaction is 1:1, the moles of nitric acid used in the titration is also 0.007 mol.
Next, let's calculate the molar concentration of the nitric acid:
molar concentration of HNO3 = moles of HNO3 / volume of HNO3 (in L)
volume of HNO3 = 14 mL = 14/1000 L = 0.014 L (convert mL to L)
molar concentration of HNO3 = 0.007 mol / 0.014 L = 0.5 mol/L
Therefore, the molar concentration of the nitric acid is 0.5 mol/L.
To find the molar concentration of the nitric acid, we can use the concept of stoichiometry and the principle of equivalent moles.
Here's how we can calculate it step by step:
1. Determine the number of moles of NaOH used:
The volume of the NaOH solution used is 28 mL, and the molar concentration is 0.25 M.
Therefore, the number of moles of NaOH can be calculated using the formula:
Moles of NaOH = Molar concentration × Volume
= 0.25 mol/L × 0.028 L
= 0.007 mol
2. Apply the principle of equivalent moles:
The balanced chemical equation between NaOH and nitric acid (HNO3) is:
NaOH + HNO3 → NaNO3 + H2O
From the balanced equation, we can see that the mole ratio between NaOH and HNO3 is 1:1.
Therefore, the number of moles of HNO3 is also 0.007 mol.
3. Determine the molar concentration of HNO3:
The volume of the HNO3 solution used is 14 mL.
Therefore, the molar concentration of HNO3 can be calculated using the formula:
Molar concentration = Moles of HNO3 / Volume
= 0.007 mol / 0.014 L
= 0.5 M
So, the molar concentration of the nitric acid is 0.5 M.