A solution of sodium hydroxide (NaOH) was standardized against potassium hydrogen phthalate (KHP). A known mass of KHP was titrated with the NaOH solution until a light pink color appeared using phenolpthalein indicator. Using the volume of NaOH required to neutralize KHP and the number of moles of KHP titrated, the concentration of the NaOH solution was calculated. Molecular formula of Potassium hydrogen phthalate: HKC8H4O4 Mass of KHP used for standardization (g) 0.5053 Volume of NaOH required to neutralize KHP (mL) 13.4473 Calculate the concentration of NaOH solution in (mol/L)

See explanation below

Explanation:

In order to calculate this, we need to use the following expression to get the concentration of the base:

MaVa = MbVb (1)

We already know the volume of NaOH used which is 13.4473 mL. We do not have the concentration of KHP, but we can use the moles. We have the mass of KHP which is 0.5053 g and the molecular formula. Let's calculate the molecular mass of KHP:

Atomic weights of the elements to be used:

K = 39.0983 g/mol;  H = 1.0078 g/mol;  C = 12.0107 g/mol;  O = 15.999 g/mol

MM KHP = (1.0078*5) + (39.0983) + (8*12.0107) + (4*15.999) = 204.2189 g/mol

Now, let's calculate the mole of KHP:

moles = 0.5053 / 204.2189 = 0.00247 moles

With the moles, we also know that:

n = M*V (2)

Replacing in (1):

n = MbVb

Now, solving for Mb:

Mb = n/Vb  (3)

Finally, replacing the data:

Mb = 0.00247 / (13.4473/1000)

Mb = 0.184 M

This would be the concentration of NaOH

pop pop pop on a opp

the answer is the first option

explanation:

just googled it m8. not that hard

lol