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star_girl

  • one year ago

Help Plz!!! 22.5cm^3 of sodium hydroxide solution reacted with 25.0cm^3 of 0.100M hydrochloric acid. Calculate the concentration of the sodium hydroxide.

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  1. star_girl
    • one year ago
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    NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l)

  2. Rushwr
    • one year ago
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    moles of NaOH : Moles of HCl is 1:1 As u can see the no. of moles HCl is equal to the no. of moles of NaOH. Using \[C =\frac{ n }{ V}\] C= concentration n= no. of moles V= volume We can find the no. of moles of HCl right? as we know the concentration and the volume of HCl . Now subject "n" in that equation n= CV Here the concentration is given in moldm^-3 So we have to convert the volume to dm^3 in order to cancel off the units. \[1cm ^{3}= 10^{-3}dm ^{3}\] \[V = 25 * 10^{-3} dm ^{3}\] therefore no. of moles of HCl ; \[n = 0.1moldm ^{-3} * 25 * 10^{-3} dm ^{3}= 2.5 * 10^{-3} mol\]We know No. of moles of HCl = no. of moles NaOH So no of moles of NaOH = 2.5 x 10^-3 mol The information we know about NaOH now is 1)the no. of moles 2) the volume Here also we need to convert the volume in to cubic decimeter. so \[V = 22.5 * 10^{-3} dm ^{3}\] Using the data we know we can find the concentration using the equation \[C = \frac{ n }{ V }\] \[C = \frac{ 2.5 * 10^{-3} mol}{ 22.5 * 10^{-3} dm ^{3} } = 0.11M\]

  3. Rushwr
    • one year ago
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    Unless there's a simplification error the answer is correct.

  4. star_girl
    • one year ago
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    thank you @Rushwr

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