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7% to 4% HCL

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  • 7% to 4% HCL

    Hi,

    What formula do I use to convert 7% HCL in 1000 mL to 4%.

    Thanks.

  • #2
    Originally posted by Alex View Post
    Hi,

    What formula do I use to convert 7% HCL in 1000 mL to 4%.

    Thanks.
    C1 * V1 = C2 * V2
    If you start with 1 L of 7% you can dilute to V1 = 0.07*1 L/0.04 = 1.75 L of 4%.

    This will require about 1.75 L - 1 L = 0.75 L of water, but always dilute to "final volume" don't measure the "added water."
    Solutions are not precisely additive when diluting high concentrations.

    Comment


    • #3
      Thanks for the reply.

      Is this correct?

      4% (required %) x 1000 mL / 7% ( original %) = 571.42 mL of water to be added.

      1000 + 571.42= 1571.42 mL.
      1571.42 -1000= 571.42 mL of water to be added.

      But that would increase the net volume more than 1000 mL?

      How do I dilute to final volume?

      Comment


      • #4
        Originally posted by Alex View Post
        Thanks for the reply.

        Is this correct?

        4% (required %) x 1000 mL / 7% ( original %) = 571.42 mL of water to be added.

        1000 + 571.42= 1571.42 mL.
        1571.42 -1000= 571.42 mL of water to be added.

        But that would increase the net volume more than 1000 mL?

        How do I dilute to final volume?
        On the first part, I am unclear what you are trying to do. Do you have 1 L of 7% and want to fully dilute it to 4%, or do you want to make 1 L of 4%.

        On the 2nd part, you need a graduated measuring jug, you measure out the required stock, then you add water to reach the specified dilute volume.

        If you want 1 L of 4% then you need 0.04*1 L/0.07 = 571.4 of your 7% stock, then dilute it by filling the measuring jug to the 1 L line.

        The final volume and the final concentration go together on one side of the equation, initial concentration and volume of stock on the other.

        Comment


        • #5
          Correct, I have 1 L of 7%, need to convert to 4% in the same volume.

          Thanks.

          Comment


          • #6
            Originally posted by JohnS View Post

            C1 * V1 = C2 * V2
            If you start with 1 L of 7% you can dilute to V1 = 0.07*1 L/0.04 = 1.75 L of 4%.

            This will require about 1.75 L - 1 L = 0.75 L of water, but always dilute to "final volume" don't measure the "added water."
            Solutions are not precisely additive when diluting high concentrations.

            I have a 35.6% (approx 36%) HCL in 2.5 liters. Want to convert it to 5%. in 125 mL.


            As per your formula C1* V1 = C2 * V2

            5 * 125 / 36 = 17.3 mL.

            I would need to add 17.3 mL of 36% HCL to 125 mL of water to be added?

            Kindly correct.


            OR


            36 * 1000/ 5 = 7,200 mL

            7,200 mL - 1000 mL = 6,200 mL or .62 L of water to be added to 125 mL?


            36 * 125 /5 = 900 mL.

            900 mL - 125 mL = 775 mL of water to be added to 125 mL?

            Thanks.
            Last edited by Alex; 07-01-2020, 02:20 AM.

            Comment


            • #7
              Originally posted by Alex View Post


              I have a 35.6% (approx 36%) HCL in 2.5 liters. Want to convert it to 5%. in 125 mL.


              As per your formula C1* V1 = C2 * V2

              5 * 125 / 36 = 17.3 mL.

              I would need to add 17.3 mL of 36% HCL to 125 mL of water to be added?

              Kindly correct.


              OR


              36 * 1000/ 5 = 7,200 mL

              7,200 mL - 1000 mL = 6,200 mL or .62 L of water to be added to 125 mL?


              36 * 125 /5 = 900 mL.

              900 mL - 125 mL = 775 mL of water to be added to 125 mL?

              Thanks.
              In the first case, add water to dilute to 125 mL final volume (about 117.3 mL of water but recall volumes are not strictly additive.

              Not sure what you are doing in the second example but 1 L = 1000 mL

              Comment


              • #8
                Originally posted by JohnS View Post

                In the first case, add water to dilute to 125 mL final volume (about 117.3 mL of water but recall volumes are not strictly additive.

                Not sure what you are doing in the second example but 1 L = 1000 mL

                For 125 mL 117.3 mL of water and 7.7 mL of 36% HCL. Correct?


                In the second example was using your earlier described method...


                C1 * V1 = C2 * V2
                If you start with 1 L of 7% you can dilute to V1 = 0.07*1 L/0.04 = 1.75 L of 4%.

                This will require about 1.75 L - 1 L = 0.75 L of water, but always dilute to "final volume" don't measure the "added water."
                Solutions are not precisely additive when diluting high concentrations.


                .36 * 1 L / .05 = 7.2 L of 5%.

                7.2 L - 1 L = 6.2 L of water to be added to 1 L of 36% HCL to make a 5%?

                Comment


                • #9
                  I have 2.5 L of 36% HCL.

                  Comment


                  • #10
                    Originally posted by Alex View Post
                    I have 2.5 L of 36% HCL.
                    But do you want to dilute it all to 5%. If so 0.36 * 2.5 L/0.05 = 18 L diluteo to 18 L final volume (requires about 15.5 L water, but measure final volume).

                    Comment


                    • #11
                      Thanks for the expert comments.

                      If I want the volumes to be the same i. e approx 100 mL on both sides.

                      36*100/5= 720 mL.

                      720-100= 620 is the volume of water needed to add to 100 mL of 36% HCL. Correct.

                      Comment

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