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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.
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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star_girl
 one year ago
Best ResponseYou've already chosen the best response.0NaOH (aq) + HCl (aq) > NaCl (aq) + H2O (l)

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1moles 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 \[C = \frac{ n }{ V }\] \[C = \frac{ 2.5 * 10^{3} mol}{ 22.5 * 10^{3} dm ^{3} } = 0.11M\]

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1Unless there's a simplification error the answer is correct.
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