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Posted by on Tuesday, February 19, 2008 at 6:48pm.

My problem is calculate the molarity of HCl and NaOH.

(molarity of acid)x(volume of acid)=(molarity of base)+(volume of added base)

HCl=acid NaOH=base

HCl volume = 25mL=0.025L
NaOH=67.05mL=0.06705

molarity = moles of solute
Liters of a solution
M=0.015000 moles of NaOH
0.06705L
M1V1=M2V2
M1V1=M2V2
V1
M1=M2V2
V1
M1=0.223713647 M of NaOH * 0.06705L
0.025L
M1=0.600000001M of HCl

0.600000001 M of HCl* 0.025L =
0.223713647 M of NaOH * 0.06705L

Am I even close?

  • chemistry - , Tuesday, February 19, 2008 at 8:25pm

    You would have done much better to type in the problem before showing your solution. It's good to show your work, however, you have no data (at least none identifiable). For example, where did the 0.01500 mols come from? Finally, you have made a typo in the second sentence (the formula). You show a + sign and it should be a x sign. If you will post the problem someone will go through and check it for you. Thanks for using Jiskha.

  • chemistry - , Tuesday, February 19, 2008 at 9:47pm

    Calculate the molarity of the HCl from the volumes of acid and base and the molarity of the NaOH. Use the following equation:

    (molarity of acid)x(volume of acid)=(molarity of base)x(volume of added base)

    equivalance point=number of mL of base to an acid

    1mL=0.001L

    HCl volume=25 mL=0.025L
    NaOH=67.05mL=0.06705L

    Molarity=Moles of solute
    ---------------
    Liters of a solution

    M=0.015000 moles of NaOH
    ----------------------
    0.06705L

    M=0.223713647M of NaOH

    M1V1=M2V2

    M1=M2V2
    ----
    V1

    M1=0.223713647M of NaOH * 0.06705L
    ---------
    0.025L

    M1= 0.600000001 M of HCl

    0.600000001 M of HCl * 0.025L=
    0.223713647 M of NaOH * 0.06705L

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