How many grams of Naoff are needed to prepare 500 ml of 0.125 M NaoH? 4. A solut

Question

How many grams of Naoff are needed to prepare 500 ml of 0.125 M NaoH? 4. A solution of malonic acid, H2C,H,O4, was standardized by titration with 0.1000 M NaOH solution. If 20.76 mL of the NaOH solution is required to neutralize completely 12.95 mL of the malonic acid solu- tion, what is the molarity of the malonic acid solution? 5. Sodium carbonate is a reagent that may be used to standardize acids in the same way you used KHP in this experiment. In such a standardization, it was found that a 0.512 g sample of sodium carbonate re- quired 26.30 mL of a sulfuric acid solution to reach the end point for the reaction. Na2co, (aq) + H2so, (aq)- H 20(1) + CO2(g) + Na2SO4(aq) yright © 2015 Pearson Education, Inc.

Explanation / Answer

3) Molarity = weight /MWtX1000/V in ml

we have to find out the weight of Sodium hydroxide MWt =40, V= 500 ml, M=0.125 M

0.125= weight/40X1000/500

weight = 2.5 grams

4) Molarity = number of molesX1000/Volume in ml

number of moles =Molarity Xvolume in ml/1000

Number of moles of NaOH = 0.1 X20.76/1000 = 0.00207 mol

number of moles of malonic acid = 0.00207 mole of NaOH X1 Mol of malonic acid /2 mol of NaOH

0.00207/2 =0.00103 mol

Molarity of malonic acid = number of moles of malonic acid X 1000/ volume in ml

= 0.00103 X1000/12.95

= 0.0795 M

5) question is not properly asked

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