9.504.930.120.500.249.2510.005.050.120.500.249.7510.505.290.240.500.4810.2511.005.440.150.500.3010.7511.505.590.150.500.3011.2512.005.800.210.500.4211.7512.505.910.110.500.2212.2513.006.070.160.500.3212.7513.506.120.050.500.1013.2514.006.230.110.500.2213.7514.506.280.050.500.1014.2515.006.360.080.500.1614.7515.506.440.080.500.1615.2516.006.480.040.500.0815.7516.506.630.150.500.3016.2517.006.780.150.500.3016.7517.506.920.140.500.2817.2518.007.140.220.500.4417.7518.507.260.120.500.2418.2519.007.380.120.500.2418.7519.507.440.060.500.1219.2520.007.510.070.500.1419.7520.507.690.180.500.3620.2521.007.820.130.500.2620.7521.507.940.120.500.2421.2522.008.020.080.500.1621.7522.508.130.110.500.2222.2523.008.190.060.500.1222.7523.508.280.090.500.1823.2524.008.350.070.500.1423.7524.508.430.080.500.1624.2525.008.520.090.500.1824.7512.500.000.000.000.000.000.00[H3PO4] in Original Cola Sample (mol/L)6.92E-03
Table 3. Determination of KaValues5 ptsKGraphical AnalysisLiterature ValuesV1/2(mL)pKaKapKaKaKa13.002.751.78E-032.17.50E-03Ka214.006.235.89E-077.26.20E-08pKa1calculated from initial pH 0.00692Plot 1. Titration Curve with ApproximateFirst Derivative0.02.04.06.08.010.012.0pHDiscussion
Discussion of StandardizationThe technical objectives of the standardization performed in this experiment was to calculate the exact molarity of the 0.01x M NaOH through standardizing NaOH using a visual endpoint acid/base titration. It was important to do a standardization by titrating a solution of KHP dissolved in water with NaOH to determine the exact concentration of NaOH to three significant figures, using the mass of KHP measured and the volume of NaOH added. The mean concentration of sodium hydroxide was calculated to be 0.010 and its standard deviation was calculated to be 0, because the calculated [NaOH] was consistent for both trials. The concentration of sodium hydroxide was calculated by converting the grams of KHP added into moles and dividing it by the volume of NaOH added: (trial 1) 0.064g KHP x1mol KHP204.2g KHPx1mol NaOH1molKHPxmol NaOH0.0317L NaOH= 0.010 M NaOH.The concentrations of [NaOH] calculated for both trials was 0.010 M, and therefore the average concentration of the two trials was 0.010 M. These concentrations can be claimed to be very precise because the standard deviation was 0, implying that there was no significant deviation between the two trials. If there were to be any variation, two likely reasons would be differences in KHP measured or the lack of precision of the buret and/or over titrating the analyte solution.