# Problem: A 25.00-mL sample of H2SO4(aq) requires 22.65 mL of the 0.5510 M NaOH for its titration. Assuming that sulfuric acid behaves as a strong acid with respect to both ionizable hydrogen atoms, what was the concentration of the sulfuric acid (H2SO4)?The Kjeldahl analysis is used to determine the protein content of foods. The food analyzed is heated with sulfuric acid in the presence of a catalyst. The carbon in the protein is converted to CO2(g), the hydrogen to water, and the nitrogen to (NH4)2SO4. The mixture is then treated with concentrated NaOH(aq) and heated to drive off NH3(g). The NH3(g) is neutralized by an excess standard acid, and the acid present after the neutralization is titrated with a standard base. From the titration data and the known mass of the sample, the percentage of N can be calculated.Based on the fact that the average percentage of nitrogen in most plant and animal protein is 16.0%, the percentage of protein is obtained by multiplying the percentage of N by the factor (100/16) = 6.25.

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A 25.00-mL sample of H2SO4(aq) requires 22.65 mL of the 0.5510 M NaOH for its titration. Assuming that sulfuric acid behaves as a strong acid with respect to both ionizable hydrogen atoms, what was the concentration of the sulfuric acid (H2SO4)?

The Kjeldahl analysis is used to determine the protein content of foods. The food analyzed is heated with sulfuric acid in the presence of a catalyst. The carbon in the protein is converted to CO2(g), the hydrogen to water, and the nitrogen to (NH4)2SO4. The mixture is then treated with concentrated NaOH(aq) and heated to drive off NH3(g). The NH3(g) is neutralized by an excess standard acid, and the acid present after the neutralization is titrated with a standard base. From the titration data and the known mass of the sample, the percentage of N can be calculated.

Based on the fact that the average percentage of nitrogen in most plant and animal protein is 16.0%, the percentage of protein is obtained by multiplying the percentage of N by the factor (100/16) = 6.25.