Titrated with standardized 01006 m using the data

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titrated with standardized (0.1006 M). Using the data from the experiment the alkalinity (0.1452 M ±0.0002), bicarbonates (0.03944M ±0.0002) and carbonates (0.106M ±0.002) content of the solution was calculated. Results and Discussion Tables Table I. Results for total Alkalinity Flas k Average HCl(M) HCl Needed (L) HCl Needed (mol) Total Alkalinity(mol) Volume of unknown (L) Total alkalinity (M) 1 0.1006 ±0.000 1 0.03608 ±0.0000 1 0.00363 0.003629648 0.025 ±0.0000 3 0.14518 6 ±0.0002 2 0.1006 ±0.000 1 0.03605 ±0.0000 1 0.003627 0.00362663 0.025 ±0.0000 3 0.14506 5 ±0.0002 Ave. 0.003628139 Ave. 0.14512 6 St. Dev 2.13405E-06 St. Dev 8.54E- 05 RSD % 0.06% RSD % 0.06% Table I. shows the results of the results of the titration to find out total alkalinity in the unknown solution (sample 13). Titration of the solution was done with standardized HCl and bromocresol blue, and this gave the content of total Carbonate and bicarbonate ions in the sample. Table II. NaOH excess used Flask Ave. NaOH(M) Total NaOH added(L) Total NaOH(mol) 1 0.1167 ± 0.0001 0.025 ± 0.00003 0.002918 ± 4.3E-06 2 0.1167 ± 0.0001 0.025 ± 0.00003 0.002918 ± 4.3E-06 Table II. Show the excess NaOH used for the back titration of the unknown solution. The NaOH react with the bicarbonate ions in the solution and this helps to calculate the bicarbonate ion content of the system by titrating the excess NaOH.
Table III. Results for the titration of excess NaOH Flask HCl(M) HCl Volume Needed (L) HCl(mol) Excess NaOH(mol) 1 0.1006 ± 0.0001 0.01893 ± 0.00001 0.001904 0.001904 ± 2.1E-06 2 0.1006 ± 0.0001 0.01947 ± 0.00001 0.001959 0.001959 ± 2.2E-06 Table III. gives the results of the titration of the excess NaOH in the system. By calculating the unreacted NaOH the amount of bicarbonate in the sample can be calculated. Table IV. Bicarbonate content in the sample Flask Total NaOH(mol) Excess NaOH(mol) NaOH reacted with HCO3 (mol) (HCO 3 ) - (mol) Volume of unknown (L) [NaHCO3] (M) 1 0.00291 8 ±4.3E -06 0.001904 ±2.14E- 06 0.001013 ±4.8E- 06 0.001013 0.025 ±0.0000 3 0.0405 ±0.0002 2 0.00291 8 ±4.3E -06 0.001959 ±2.19E- 06 0.000959 ±4.8E- 06 0.000959 0.025 ±0.0000 3 0.0383 ±0.0002 Ave. 0.0009859 8 Ave. 0.03943 9 St. Dev. 3.84129E- 05 St. Dev. 0.00153 7 RSD % 3.90 RSD % 3.90 Table IV. provide the carbonate content of the unknown solution (sample 13). Using the data from the back titration of the NaOH the bicarbonate content can be found and through that the carbonate content could be deduced. Table V. Carbonates in the sample Average total alkalinity (mol) Average (HCO 3 ) - (mol) Average (CO 3 ) -2 (mol) Volume of unknown(L) [Na2CO3] (M) X ±S 0.003628 ±2.13E-06 0.00099 ±3.84E- 05 0.002642 0.025 ±0.00003 0.106 ±0.002 Table V. indicate the carbonate content in the unknown solution (sample 13). Total alkalinity and bicarbonate amount was used to determine the carbonate content in the solution. Sample calculations In the experimental procedure titration was carried out using standardized HCl and the unknown solution (sample 13) to determine the total alkalinity of the sample. Table I depicts the data of that titration. Molesof HClneeded = Molarityof HCl Volumeof HCl used [4] Equation [4] is used to calculate how many moles of HCl was used to neutralize the carbonates in the sample. The HCl needed for the titration of the flask 1 of the table I can be calculated as below. Molesof HClneeded = 0.1006 mol L 0.03608 L = 0.003630 mol
In the titration for the table I, moles of HCl reacted is related to the carbonates and bicarbonates in the solution by the equation [5]. 2 ¿ = Total Alkalinity ¿ + Molesof CO 3 ¿ Molesof HCl = Molesof HCO 3 ¿ [5]

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