Question

Can you do number 3

Construct and calculate a theoretical titration curve, showing and discussing all key areas, for the titration of 0.100 M NaOH with 50.0 mL 0.100 M HNO_2 (K_aHNO_2 = 7.1 times 10^-4) A 30.0 L of air sample (Density of 1.20 g/L) was passed through an absorption tower containing a solution of Cd^2+, where H_2S was retained as CdS. The mixture was acidified and treated with 10.0 mL of 0.01070 M I_2. After the reaction S^2-_ + I_2 - S(s) + 2I^- was complete, the excess idodine was titrated with 12.85 mL of 0.01344 M thiosulfate, S_2O_3^2-. Calculate the H_2S in ppm. Mol. Mass H_2S = 34 g/mol A 6.881-g sample containing MgCl_2 and NaCl was dissolved in sufficient water to give a 500-mL volume. Analysis of the chloride content in a 50-0mL aliquot resulted in the formation of 0.5923-g of AgCl. The magnesium in al second 50-0 mL aliquot was precipitated as MgNH_4PO_4; on ignition 0.1796-g of Mg_2P_2O_7 was found. Calculate the percentages of MgCl_2.H_2O and NaCl in the original sample. Mol. Mass: MgP_2O_7 = 222.55 g/mol, NaCl = 58.44 g/mol, MgCl_2.6H_2O = 203.30 g/mol The homogeneity of a standard chloride solution was tested by analyzing portions of the material at the top and bottom with the following found Is homogeneity indicated at the 95% confidence level, use both t-test and f-test. DERIVE and calculate the solution potential for the titration of Sn^2+ with Cr_2O_7^2- at The reaction: 3Sn^2+ + Cr_2O_7^2- + 14H^+ = 2Cr^3+ 3 Sn^4+ + 7 H_2O

Answer 1

Answer (3)

• First of all 6.881 g of MgCl2.6H2O and NaCl dissolved in 500 mL. One tenth of that (50 mL) gave 0.5923 g of AgCl (mol. wt. = 143.4 g) so 0.5923/143.4 = 0.004132 mole AgCl and the same number of moles of Cl.
• Mg2P2O7 has mol. wt. 222.6 g so 0.1796/222.6 = 0.0008068 moles of Mg2P2O7 which means 0.001614 mole Mg (two mole Mg per mol Mg2P2O7)
• So 500 mL contained 10x that moles Mg = 0.01614 which means that solid contained 0.01614 moles MgCl2.6H2O (mol. wt. = 203.3) so 0.01614*203.3 = 3.281 g which leaves 3.600 g of NaCl.
• 3.600 g NaCl are (3.600/58.5) moles = 0.06160 moles of NaCl which means the sample contained 0.0616 +( 2 x 0.01614) moles Cl = 0.09387 moles which is not the answer for the moles of Cl.
• If the former, moles Cl in NaCl = 0.4132 - 2*0.01614 = 0.009044 moles so g NaCl = 0.009044*58.5 = 0.5287 g.
• That suggests 3.281 g are magnesium chloride; 0.5287 are NaCl. %ages are 47.68%and 7.68%