Titration experiment, measuring the molarity of Sodium Hydroxide

Measuring the molarity of Sodium Hydroxide(NaOH) with a reaction with a known quantity of strong acid - Hydrochloric Acid


In this experiment, measuring the molarity of the provided sodium hydroxide solution by using the given 31.45% Hydrochloric acid.

Difficulty: Easy , but need adult supervision and lots of care because of the chemicals


  1. Sodium Hydroxide solution: 75 ml

  2. Hydrochloric acid (31.45% solution): 100 ml

  3. Calibrated Measuring device

  4. Turmeric powder as ph indicator


  1. Add ½ a teaspoon of turmeric powder to the provided Sodium Hydroxide solution. The solution will turn red

  2. Carefully add the hydrochloric acid solution till the solution turns yellow.

  3. Measure the amount of hydrochloric acid added at that time. With this, we can calculate the molarity of the provided HCL as shown below. 12 ml of HCL was required to neutralize the base

How It Works:

Hydrochloric acid is a strong acid with a pH value close to 2. Sodium Hydroxide is a very strong base with a value of 14. They react per the equation below generating water and Sodium Chloride (Common Salt) per equation beow

NaOH (aq) + HCl (aq) → H2O (aq) + NaCl (aq)

Turmeric is a pH indicator that will be red in bases and yellow in neutral and acidic solutions. Once the turmeric turns yellow, it means that the base (sodium hydroxide) has been neutralized.

Calculation of molarity :

1. Molarity of 31.45% HCl is 9.995 mol/liter

2. The number of moles in 12 ml of HCL (0.012 L) is 9.995 * 0.012 = 0.112 mol of HCl

3. The reaction equation is

HCl (aq) + NaOH (aq) —> H2O (l) + NaCl (aq)

4. For every 1 mol of NaOH used, 1 mol of HCl is used

5. So 0.112 mol of NaOH is used

6. The molar mass of NaOH = (22.99)+16+1.008 = 39.998 g/mol

7. The weight of NaOH in the solution is 39.998 * 0.112 = 4.48 g

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