uppose you are standardizing a sodium hydroxide solution with K H P (molar mass=204.2 g/mol) according to the equation K H P + N a O H ⟶ H 2 O + N a K P You prepare the standard solution from 0.294 g of K H P in 250.0 mL of water. You then require 7.42 mL of N a O H solution to complete the titration. What is the concentration of the N a O H solution?

The concentration of the NaOH solution used in complete titration has been 0.194 M.

The reaction of KHP with NaOH has been the neutralization reaction that results in the formation of salt and water.

The concentration of KHP solution has been:

Molarity =

Molarity of KHP =

Molarity of KHP = 0.0057 M

The concentration of the KHP solution has been 0.0057 M.

For the neutralization reaction in complete titration:

Molarity of acid Volume of acid = Molarity of base Volume of base

Substituting the values from the question:

0.0057 M 250 ml = Molarity of NaOH 7.42 ml

1.439 = Molarity of NaOH 7.42 ml

Molarity of NaOH = 0.194 M.

The concentration of the NaOH solution used in complete titration has been 0.194 M.

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The concentration of NaOH solution is 0.194 M

Explanation:

To calculate the number of moles, we use the equation:

Given mass of KHP = 0.294 g

Molar mass of KHP = 204.22 g/mol

Putting values in above equation, we get:

The chemical reaction follows the equation:

By Stoichiometry of the reaction:

1 mole of KHP reacts with 1 mole of NaOH.

So, 0.00144 moles of KHP will react with = 0.0144 of KOH.

To calculate the molarity of NaOH, we use the equation:

We are given:

Moles of KOH = 0.00144 moles

Volume of solution = 7.42ml  = 0.00742L      (Conversion factor:  1L = 1000 mL)

Putting values in above equation, we get:

Hence, the molarity of NaOH solution is 0.194 M

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explanation:

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