Can I please get some help with questions # 3 through # 8. Thank you! Font Famil

Question

Can I please get some help with questions # 3 through # 8. Thank you!

Font Family Font Size Notes Portfolio Graphing Data base(NaOH): acid(HCL)1:1 ratio So at the equivalence point, 50mL - 38.79mL 11.21 mL 2. Initial buret reading 50mL - 37.41mL 12.59mL 3. Volume of HCl solution used n(NaOH) n(HCL) Now you can calculate the unknown concentration of th HCl according to: C = n ÷ V or C(acid) x V(acid) = M(base) V(base) The only unknown in the above equation is the concentration of the acid which can now be found as follows: 4. Molarity of NaOH solution used Calculations - for each of following do the necessary calculation to obtain the item requested. Place your calculated result to the right of the item. Be sure to include the proper units. In the space below the item describe how you calculated your answer. (see end of experiment 3 assignment or suggestions on how to describe a calculation C(acid)-(C(base) x V(base)) / V(acid) result to the space bel e the emdescribehude he proper Sodium Hydroxide = 0.100M Hydrocholoric Acid-? add 50mL NaOH (luquid) to Buretter then add 10mL H20 into flask volume to use when calcualting concentration of HCl is Volume of NaOH used 6. Molarity of NaOH solution (M), Use Cacid x Vacid = Cbase x Vbase 10mL 2 drops of pheolphthalein dispensed 12.59 mL bright purple 50-37.41 12.59 add 10mL HCI, 10mL h20 and 2 drops, adding by drops What is the equivalence point of a titration? this time 11.21 when it turned purple 8. How does the equivalence point of a titration differ from the end point? 50mL-38. 79mL 1 1.21mL of sodium hydroxide was delivered to complete the titration

Explanation / Answer

So here we have,

A 11.21 mL of 0.1 M NaOH was used for the titration of a 10 mL HCl solution in 10 mL H2O

We know,

1 mole of NaOH would neutralize 1 mole of HCl

moles of NaOH = M x L = 0.1 M x 0.05 L = 0.005 moles = moles of HCl

thus, Molarity of HCl in 10 mL = 0.005/0.01 = 0.05 M

3) Volume of HCl used = 10 mL

4) Molarity of NaOH used for titration = 0.1 M

5) Volume of NaOH used = 11.21 mL

6) Molarity of NaOH in solution = 0.05 M of acid x 0.010 L of acid / 0.01121 L of NaOH = 0.0446 M

7) The equivalence point in titration is when amount of base equals to the amount of acid added in solution in stoichiometry.

8) End point is when the indicator shows a definite color change.   

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