Calculations for titration.
Let’s learn calculations for titration. The most accurate or helpful solution is served by ausetute.com.au.
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Best solution
Arrhenius Acid-Base Direct Titration Calculations: Key Concepts. An acid-base titration is used to determine the concentration of an acid or a base.
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Other solutions
given 0.2038 gram sample of Ni(NH3)5CL2 dissolved in 0.050L of 0.1503M HCL. Calculate how much 0.09901M NaOH would be required to titrate this sample to a endpoint ? so i know that the HCL in excess converts all the Nh3 in the sample into NH4. so i think...
Answer:
you know what to do .. here is the solution .. Molar mass of Ni(NH3)5Cl2 is 214.7 g/mole so we have...
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Answer:
A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M NH3. The Kb of NH3 is 1.8e-5. a. Determine the...
Bob at Yahoo! Answers Mark as irrelevant Undo
I have to calculate: 1. moles of NaOH 2. Concentration of NaOH Only problem is, the only data I have is (volume NaOH =27.9 mL). How do I calculate the number or moles if I only have the volume?
Answer:
Hmmmm there has to be more to this problem. I am assuming you are in high school and solving this isnt...
Ilovemus... at Yahoo! Answers Mark as irrelevant Undo
I have several more problems to do like this, but if someone could please walk me through this step by step? It is a little pressing, as I need it by tomorrow morning, but please do what you can! Thank you. ------------- Potassium dichromate reacts with...
Answer:
3 moles of potassium oxalate reacts with 1 mole of potassium dichromate. 0.2016 g K2C2O4 / 166.2 g/mol...
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Answer:
Titration is a common laboratory method of quantitative/chemical analysis which can be used to determine...
twentyfo... at Yahoo! Answers Mark as irrelevant Undo
Heres what we did: Provided with a standard 0.01M solution of sodium carbonate Exactly 25.0cm3 of the approximate 2M hydrochloric acid was very carefully diluted using a bulb pipette. The meniscus was exactly on the line of 25.0cm3 on the pipette. The...
Answer:
moles of Na2CO3 = M x V = .01 mole liter ^-1 x .025 liter = 0.00025 moles we see from the equation that...
Bobby_Th... at Yahoo! Answers Mark as irrelevant Undo
scenario: - 50mL vinegar is diluted to 500mL - 25mL of this diluted vinegar is titrated with 0.1M 26.167mL NaOH solution What is the concentration of acetic acid in this vinegar? ---- I know that no. of NaOH moles = 2.6167x10^-3 moles = no. of moles...
Answer:
NaOH reacts with CH3COOH in 1:1 molar ratio Mol NaOH in 26.167mL of 0.1M solution = 26.167/1000*0.1...
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I have done a Mohr titration in 2nd year chem and struggling with the calculations, can someone help please. Have asked my lecturer but been led astray!! Here are the details: Started with an unknown sample (containing .22-.44g of KCl and .50-1.00g KI...
Answer:
There are several mistakes in the calculation of KI. You stated that the AgNO3 has a concentration of...
cbr600_g... at Yahoo! Answers Mark as irrelevant Undo
Please help on my titration lab, I've worked on it for a while and it's due later today!! The following list is what I have to do: Moles of NaOH which reacted: Moles of tartaric acid which reacted with OH^-(mol) Mass of tartaric acid present in weighed...
Answer:
OK here is the balanced equation C4H6O6 + 2NaOH -> Na2C4H4O6 + 2H2O moles of NaOH at the equivalence...
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