10.0 ml of 0.05 M KMnO₄ solution was consumed in a titration with 10.0 mL of given oxalic acid dehydrate solution. The strength of given oxalic acid solution is _____ ×10⁻² g/L. (Round off to the Nearest Integer).
10.0 ml of 0.05 M KMnO₄ solution was consumed in a titration with 10.0 mL of given oxalic acid dehydrate solution. The strength of given oxalic acid solution is _____ ×10⁻² g/L. (Round off to the Nearest Integer).
-
1 Answer
-
MnO? + H? C? O?2H? O - (H? )-> Mn²? + CO?
nf = (5) nf = (2)
Milli equivalent of C? O? ²? = mili equivalent of MnO?
2 x M x 10 = 5 x 0.05 x 10
M = 0.125 M
M = Strength / M.M of H? C? O?2H? O
Strength = 0.125 x 126g/L
= 15.75g/L
Similar Questions for you
After balancing change in oxidation state,
2Cl2 2Cl– + 2ClO–
Next, balance 'O' atoms,
2Cl2 +4OH–
Simplifying to get the simplest ratios,
Cl2 +2OH–
x = 1, y = 2, z = 1, p = 1
Disproportionation reaction is a reaction in which a substance (element) is simultaneously oxidised and reduced.
Potassium hydrogen phthalate is used to standardize NaOH solutions.
Phenolphthalein is used as an indicator to detect completion of titrations.
meq. of K2Cr2O7 = meq. of NO–2
n1 * 6 = 1 * 2
n1=1/3
Oxidation no. of a substance is proportional to the charge on it. In this case the charge is +
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 65k Colleges
- 1.2k Exams
- 687k Reviews
- 1800k Answers