Jodate: IO3- + 6e- + 6H+ --> I- + 3H2O
equivalency to:
Thio: 6S2O32- --> 3S4O62- + 6e-
25.0 ml * 0.106 mmol/ml (KIO3) = 2.65 mmol KIO3 reacting (1:6) with Thio: 6*2.65 = 15.9 mmol sodium thiosulfate. This is in 16.25 ml, so the molarity is 15.9 mmol / 16.25 =
= 0.978 M thiosulfate
it will increase the molarity of the acetic acid
I need to see the balanced equation to work!NaOH + HCl --> NaCl + H2O ( good, all one to one )Now, find molarity HCl ( sodium, or sodium hydroxide; no matter )(17.65 mL)(0.110 M NaOH) = (25.00 mL)(X M HCl)= 0.07766 M HCl-------------------------now,Molarity = moles of solute/Liters of solution ( 25.00 mL = 0.025 L)0.07766 M HCl = X moles/0.025 Liters= 0.001942 moles HCl---------------------------------------formal set up, though not needed0.001942 moles HCl (1 mole NaOH/1 mole HCl)= 0.00194 moles sodium hydroxide used=============================
Mass of H2O2 = 0.637 g
Sodium hydroxide cannot be used as a primary standard because of its hygroscopic properties as a solid. Because it is so prone to absorbing water, it is impossible to accurately measure the mass of a solid sample, so instead it must be put into solution and titrated with a known acidic solution. This makes it a secondary standard as opposed to a primary standard. The same logic holds true for hydrated sodium carbonate. Any solution that MUST be titrated in order to know it true molarity is considered a secondary standard. It also reacts with atmospheric carbon dioxide to form sodium carbonate, so it is unstable.
Well the question is not very specific, but I think you may be reffering to the boiling of water before acid-base titrations. Water is boiled previous to acid-base titrations because if left open to the atmosphere @ STP water will absorb carbon dioxide until the pH is roughly 5.5. This will obviously adversly affect the accuracy of an acid-base titration. In the permanganate titration, a redox reaction occurs in the presence of an acidic solution.
Completely titrated means it reached the stoichiometric point (usually pH=7). Simply means neutralized.
0.289 Moles is the molarity of an NaOH solution if 4.37 ml is titrated by 11.1 ml of 0.0904m hno3.
The HCL concentration is 1.2M or 1.2N
0.114 M
In iodometric titrations sodium thiosulfate is the titrant whereas the KI will reduce the analyte; eg: Cu2+ to Cu+. The I2 produced is then titrated by the sodium thiosulphate. Cu2+ + I- --> CuI + I3- I3- + 2 S2O32- ¾® 3 I- + S4O62- To answer your question: KI (reducing agent) is added to generate the iodine by the reduction of the analyte (Cu2+) The formed iodine is then back-titrated with thiosulfate (titrant) to determine the amount of analyte originally present. As you can see the KI and sodium thiosulfate serve two different purposes. KI improves solubility of Iodine
it will increase the molarity of the acetic acid
M = n ÷ V M-Molarity (mol/l) n- moles V- volume in liters. Volume cannot be expressed in grams....
Standardization of sodium thiosulfate uses potassium iodate with excess potassium iodide and acidified. Iodine is liberated and that is titrated with sodium thiosulfate. KIO3 + 5KI + 3H2SO4 -----> 3K2SO4 + 3H2O + 3 I2 I2 + 2Na2S2O3 -------> 2NaI + Na2S4O6 So 1 mole of KIO3 produces 3 moles of Iodine. 1 moles of iodine reacts with 2 moles of thiosulfate. So 6 moles of sodium thiosulfate react with 1 mole of potassium iodate KIO3.
Cations can be titrated.
Thiosulfate: 2 S2O32- --> S4O62- + 2e-equivalency to:Chlorine: 1 Cl2 + 2e- --> 2Cl-31.6 ml * 0.141 mmol/ml S2O32- = 4.456 mmol S2O32-= 4.456 *(2 electron / 2 S2O32-) = 4.456 mmol (electrons) == 4.456 *(1 Cl2 / 2 electron) = 2.228 mmol Cl2 == 2.228 * 70.90 mg/mmol Cl2 = 158 mg == 0.158 g Chlorine
when magnesium sulfate and potassium hydroxide is added,magnesium hydroxide is formed and precipitates and conc H2SO4 is added and in presence of oxygen the preciptate turns brown color which can be measured by idoine and starch indicator titrated against sodium thiosulfate
No; acids can be titrated with bases.