5. What is the concentration of a sulfuric acid solution (a diprotic acid) which requires 37.18 mL of \( 0.0950 \mathrm{M} \mathrm{NaOH}(\mathrm{aq}) \) solution to titrate 25.00 mL of the acidic solution? Chemical equation: \( \mathrm{H}_{2} \mathrm{SO}_{4}+2 \mathrm{NaOl} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}- \) Calculation: 6. During a titration, 10.00 mL of an acidic solution of unknown concentration was placed in a 125 mL Erlenmeyer flask. To this was added 20.0 mL of deionized water and two drops of

To solve for the concentration of the sulfuric acid, we start by using the stoichiometry of the reaction. The balanced equation shows that two moles of NaOH react with one mole of sulfuric acid. Given that 37.18 mL of 0.0950 M NaOH was used, we find the moles of NaOH: Moles of NaOH = \(0.0950 \, \text{mol/L} \times \frac{37.18 \, \text{mL}}{1000 \, \text{mL/L}} = 0.00353 \, \text{mol}\). Thus, the moles of sulfuric acid (since it's a diprotic acid) are half that: Moles of \(\mathrm{H}_{2}\mathrm{SO}_{4} = \frac{0.00353}{2} = 0.001765 \, \text{mol}\). To find the concentration of the sulfuric acid solution, use the formula: Concentration = \(\frac{\text{moles}}{\text{volume in Liters}} = \frac{0.001765}{0.025} = 0.0706 \, \mathrm{M}\). Tada! The concentration of the sulfuric acid solution is \(0.0706 \, \mathrm{M}\). As for the second part of your query on titrating the acidic solution, it’s interesting to note that diluting the solution before titration can impact the results if you're not careful. Although adding deionized water helps in mixing, it also dilutes the acid, potentially leading to confusion if calculations are based on the original concentration. Remember, dilution doesn’t affect the number of moles of acid present; it just changes the concentration. Always keep the final total volume in mind during calculations!