Question

Two aqueous solutions are mixed: \(50.0 \mathrm{mL}\) of 0.0150 \(\mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) and \(50.0 \mathrm{mL}\) of \(0.0385 \mathrm{M} \mathrm{NaOH} .\) What is the pH of the resulting solution?

Short answer

The pH of the resulting solution is 11.37

Step-by-step solution

Analysis of the Reaction

Reacting sulphuric acid and sodium hydroxide, we get water and sodium sulphate. The reaction is given as: \(H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O\). This is a 1:2 reaction.

Calculating Moles

Here, volume is in ml but for our calculations, we want it in liters. So, 50 ml is converted to liters by dividing with 1000 giving \(0.050 L\). Then, moles of sulphuric acid and sodium hydroxide are calculated by multiplying molarity with volume: \[n= MV \Rightarrow n(H_2SO_4)=0.015 * 0.050 = 0.00075 mol \] \[n= MV \Rightarrow n(NaOH) = 0.0385 * 0.050 = 0.001925 mol\]

Determining the Limiting Reactant and Excess Base

In this reaction, double moles of \(NaOH\) are needed. But, the number of moles of \(NaOH\) is not twice that of \(H_2SO_4\), so \(H_2SO_4\) is the limiting reactant and all of it is used. Then, calculate the remaining \(NaOH\), subtract the used \(NaOH\) from the total: \[n_{NaOH(left)} = n_{NaOH(total)} - 2 * n_{H_2SO_4} = 0.001925 - 2*0.00075 = 0.000425 mol\]

Calculate the new [NaOH]

The new concentration of \(NaOH\) ('[NaOH]') is determined by taking the moles remaining and dividing by the total volume of solution (100 mL or 0.10 L): \[[NaOH] = n_{NaOH(left)}/ V_{total} = 0.000425/0.100 = 0.00425 M\]

Calculating the pH

Since sodium hydroxide is a strong base, its pOH equals the negative logarithm of [NaOH], and, hence, pH is given by subtracting the pOH from 14: \[pOH=-log[OH-]= -log[NaOH] = -log[0.00425] ; pH = 14 - pOH\]

Key concepts

Acid-Base Titration

Acid-base titration is a process used to determine the concentration of an acidic or a basic solution by reacting it with a base or an acid of known concentration until neutralization occurs. The point at which the number of moles of hydrogen ions equals the number of moles of hydroxide ions is called the equivalence point. In a titration, indicators or pH meters can be used to know when the equivalence point is reached.

For instance, in the exercise provided, sulphuric acid (\( H_2SO_4 \)) reacts with sodium hydroxide (\( NaOH \)), a typical acid-base reaction. By monitoring the volume and concentration of reactants, one can determine the pH of the resulting solution at the end of the titration. Understanding this concept is crucial as it forms the basis of calculating pH in this kind of analytical procedure.

Stoichiometry

Stoichiometry is the study of the quantitative relationships, or ratios, between the reactants and products in chemical reactions. It relies on the balanced chemical equation and the mole concept to make predictions about the quantities of substances consumed and produced in reactions.

For example, the balanced equation for the reaction between sulphuric acid and sodium hydroxide is: \( H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O \). This equation indicates that for every mole of \( H_2SO_4 \), two moles of \( NaOH \) are needed for complete reaction. Applying stoichiometry helps to determine the amount of each reactant needed and, as shown in the solution, to find the limiting reactant.

Limiting Reactant

The limiting reactant in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete. Once the limiting reactant is used up, no further reaction can occur. Identifying the limiting reactant is a critical step in stoichiometry because it determines the maximum amount of product that can be formed.

In the given exercise, sulphuric acid (\( H_2SO_4 \) is the limiting reactant as it is consumed first, limiting the amount of product formed. Despite having more moles of \( NaOH \), once all the \( H_2SO_4 \) is reacted, the excess \( NaOH \) does not have more acid to react with, and thus the remaining \( NaOH \) dictates the pH of the solution.

Molarity

Molarity, denoted as M, is a measure of concentration that expresses the moles of a solute contained in one liter of solution. It is one of the most common units used in chemistry for quantifying the concentration of substances. To calculate molarity, one can divide the number of moles of solute by the volume of solution in liters.

In the provided exercise, the molarity of \( H_2SO_4 \) and \( NaOH \) initially guides the stoichiometric calculations to establish the moles present. After determining the limiting reactant, the leftover \( NaOH \) is used to calculate the molarity of the remaining base in the solution. This final concentration is then utilized to obtain the pH, illustrating the importance of understanding molarity when solving pH-related problems.