Skip to content

    Press Enter for all results · Esc to close

    TO FIND THE EQUIVALENT WEIGHT OF GIVEN METAL VOLUMETRICALLY

    2 min read
    On this page
    1. Experiment 20: TO FIND THE EQUIVALENT WEIGHT OF GIVEN METAL VOLUMETRICALLY
    2. APPARATUS REQUIRED
    3. CHEMICALS REQUIRED
    4. THEORY
    5. PROCEDURE
    6. OBSERVATION
    7. OBSERVATION TABLE
    8. RESULT
    9. CONCLUSION
    10. PRECAUTIONS

    Experiment 20: TO FIND THE EQUIVALENT WEIGHT OF GIVEN METAL VOLUMETRICALLY

    APPARATUS REQUIRED

    • Beaker
    • Funnel
    • Volumetric flask
    • Pipette
    • Burette
    • Conical flask
    • Chemical balance

    CHEMICALS REQUIRED

    • Given metal
    • 0.2N NaOH
    • 0.5 N HCl

    THEORY

    A definite weight of metal is weighed with the help of chemical balance. It is reacted completely with a known excess volume of standard 0.5N HCl. The metal consumes a fraction of the acid. The amount of unused acid left after the reaction is determined by back titration against standard 0.2N NaOH solution. Then from the volume of the acid used by the metal, the equivalent weight of metal is calculated.

     \begin{align*}
    Mg + 2HCl  &\to MgCl_{2} + H_{2}\\ 
    HCl + NaOH &\to NaCl + H_{2}O
    \end{align*}

    According to the normality equation,

    \left ( \frac{W}{E} \right )_{metal}+\left ( \frac{V\times N}{1000} \right )_{base} = \left ( \frac{V\times N}{1000} \right )_{acid}

    PROCEDURE

    At first, exactly 0.018 gm of pure metal was weighed. 10 ml of 0.5N HCl was kept in a conical flask and the weighed metal was kept in it. It was left for the completion of the reaction and a few drops of phenolphthalein were kept in it. Now, the burette was rinsed with NaOH solution and fitted up to zero mark. Then this acidic solution was titrated with the standard NaOH solution to get the end point.

    OBSERVATION

    Weight of metal (w) = 0.018 gm

    OBSERVATION TABLE

    S.NVolume of HCl (ml)Burette Readings: NaOH (ml)
    InitialFinalDifference
    1.10017.517.5

    Calculations:

    \begin{align*} 
    \left ( \frac{W}{E} \right )_{metal}+\left ( \frac{V\times N}{1000} \right )_{base} &= \left ( \frac{V\times N}{1000} \right )_{acid}\\ 
    \left ( \frac{0.018}{E} \right )+\left ( \frac{17.5\times 0.2}{1000} \right )&= \left ( \frac{10\times 0.5}{1000} \right )\\ 
    \left ( \frac{0.018}{E} \right )+0.0035 = &0.005\\ 
    \left ( \frac{0.018}{E} \right )= 0.0015\\ 
    \therefore E = 12
    
    \end{align*}

    RESULT

    Thus, the equivalent weight of a given metal was found to be 12.

    CONCLUSION

    The equivalent weight of metal can be determined volumetrically.

    PRECAUTIONS

    1. The pipette and burette should be rinsed with corresponding solutions and the conical flask should never be rinsed.
    2. The metal should react completely with the acid.
    3. The lower meniscus of the colourless solution should be noted while taking the reading.
    4. Chemicals and glassware should be handled with care.

    Questions & discussion

    Have a question, or found this note helpful? Be the first to comment.

    Ask a question or share feedback

    Your email is never shown publicly.