What is the molarity of a NaOH solution if 26.55 ml are needed to titrate a .6939g sample of KHP?

Question

Wyatt Atkins · Accepted Answer

Approx. #0.13*mol*L^-1#............

We need a stoichiometric equation........ #underbrace(C_6H_4(CO_2^(-)K^+)(CO_2H))_"potassium hydrogen phthalate, KHP" + NaOHrarrC_6H_4(CO_2^(-)K^+)(CO_2^(-)Na^+) + H_2O# There is thus 1:1 stoichiometry........ #"Moles of KHP"=(0.6939*g)/(204.22*g*mol^-1)=3.398xx10^-3*mol.# And since this molar quantity reacted with #26.55*mL# of titrant... #"Concentration"=(3.398xx10^-3*mol)/(26.55*mLxx10^-3*L*mL^-1)# #=0.1280*mol*L^-1#.........

Avery Adams · Answer

undefined To find the molarity of the NaOH solution, we can use the concept of titration and stoichiometry.

The balanced chemical equation for the reaction between NaOH and KHP (potassium hydrogen phthalate) is:

$ \text{NaOH} + \text{KHP} \rightarrow \text{NaKP} + \text{H}_2\text{O} $

From the equation, we see that one mole of NaOH reacts with one mole of KHP.

1. Calculate the moles of KHP:
   $ \text{moles}_{\text{KHP}} = \frac{\text{mass}_{\text{KHP}}}{\text{molar mass}_{\text{KHP}}} $

2. Use stoichiometry to find the moles of NaOH:
   Since the molar ratio of NaOH to KHP is 1:1, the moles of NaOH will be the same as the moles of KHP.

3. Calculate the molarity of NaOH:
   $ \text{Molarity} = \frac{\text{moles of NaOH}}{\text{volume of NaOH solution (in liters)}} $

Given:
- Volume of NaOH solution used for titration = 26.55 mL = 0.02655 L
- Mass of KHP = 0.6939 g
- Molar mass of KHP = $204.22 \, \text{g/mol}$

Let's calculate:

1. $ \text{moles}_{\text{KHP}} = \frac{0.6939 \, \text{g}}{204.22 \, \text{g/mol}} = 0.003397 \, \text{mol} $

2. Moles of NaOH = Moles of KHP = 0.003397 mol

3. $ \text{Molarity} = \frac{0.003397 \, \text{mol}}{0.02655 \, \text{L}} = 0.1279 \, \text{M} $

Therefore, the molarity of the NaOH solution is approximately 0.1279 M.