Advance study assignments Equivalent Mass of and Unknown Acid 7.0 mL of 6.0 M Na

Question

Advance study assignments Equivalent Mass of and Unknown Acid 7.0 mL of 6.0 M NaOH are diluted with water to a volume of 400 mL. You are asked to the molarity Of the resulting solution. First find out how many moles of NaOH there are in 7.0 mL of 6.0 M NaOH. Use Equation 1. Note that the volume must be in liters. Moles b. Since the total number of moles of NaOH is not changed on dilution, the molarity after dilution can also be found by Equation 1, using the final volume of the solution. Calculate that molarity. M In an acid-base titration, 24.88 mL of an NaOH solution are needed to neutralize 26.43 mL of a 0.1049 M HCI solution. To find the molarity of the NaOH solution, we can use the following procedure: First note the value of MH+ in the HC1 solution. M Find MOH- in NaOH solution. (Use Eq. 3.) M Obtain MNAOH from MOH-. A 0.2349 g sample of an unknown acid requires 33.66 mL of 0.1086 M NAOH for neutralization to a phenolphthalein end point There are 0.42 mL of 0.1049 M HCI used of back-titration. How many moles of OH- are used? How many moles of H+from HC1 moles OH- moles OH+ How many moles of H+ are there in the solid add? (Use Eq. 6.) moles H+ in solid What is the equivalent mass of the unknown acid? (Use Eq. 4.) g

Explanation / Answer

1.

a) given:- 7 ml of 6M NaOH
   molarity=no. of moles/volume of solution in litres

so 6=n/0.007 (7ml=0.007l)

so n=0.042 moles

b) after dilution
Molarity=n/volume of solution in litres

so M=0.042/0.4 (400ml=0.4 litres)

so new Molarity = 0.105 M

2.

HCl is strong acid so it dissociates completely
so MH+ = no. of moles of H+/volume of solution in litres

MH+= 0.1049 (1 HCl molecule gives 1 H+ so Molarity will be same)

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