A student added 50cm^3 of 1.00 moldm^-3 sulphuric acid to 25cm^3 of 1.00 moldm^-3 sodium hydroxide. Calculate the pH of the solution formed.
To calculate the pH of the solution formed, we need to determine the concentrations of the resulting solution's components after the reaction between sulphuric acid (H2SO4) and sodium hydroxide (NaOH).
First, let's write the balanced chemical equation for the reaction:
H2SO4 + 2NaOH → Na2SO4 + 2H2O
From the balanced equation, we can see that the ratio of H2SO4 to NaOH is 1:2. This means that one mole of sulphuric acid reacts with two moles of sodium hydroxide to form one mole of sodium sulfate and two moles of water.
Given:
Volume of H2SO4 = 50 cm^3 = 0.05 dm^3
Concentration of H2SO4 = 1.00 moldm^-3
From the given concentration and volume, we can calculate the number of moles of H2SO4:
Moles of H2SO4 = Concentration × Volume
= 1.00 mol/dm^3 × 0.05 dm^3
= 0.05 moles
Since one mole of sulphuric acid reacts with two moles of sodium hydroxide, we can conclude that there are 0.10 moles of NaOH in the reaction.
Now, let's calculate the final volume of the solution after mixing the two solutions:
Volume of NaOH = 25 cm^3 = 0.025 dm^3
To find the concentration of the final solution, we can add the moles of H2SO4 and NaOH and divide it by the final volume of the solution:
Concentration of final solution = (moles of H2SO4 + moles of NaOH) / volume of final solution
= (0.05 + 0.10) moles / (0.05 + 0.025) dm^3
= 0.15 mol / 0.075 dm^3
= 2.00 moldm^-3
Since we have the concentration of the final solution and we know that pH is given by the formula pH = -log[H+], we can calculate the pH.
[H+] is the concentration of hydrogen ions in the solution. In our case, the concentration of hydrogen ions is equal to the concentration of sulphuric acid because it is a strong acid that fully dissociates in water.
Thus, the pH of the solution formed is pH = -log[H+]. Plugging in the concentration of the final solution, we have:
pH = -log[2.00] ≈ -0.3010 ≈ 0.301
Therefore, the pH of the solution formed is approximately 0.301.