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    25 cm^3 of 0.02 moldm^-3 of HCl was mixed with 75 cm^3 of water. Determine the pH of the diluted acid.

    the answer is 2.31 but how do i get to this?
    Thanks
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    (Original post by master y)
    25 cm^3 of 0.02 moldm^-3 of HCl was mixed with 75 cm^3 of water. Determine the pH of the diluted acid.

    the answer is 2.31 but how do i get to this?
    Thanks
    You can't -log(0.02) because the HCl concentration is more dilute when you add the water.

    Because of this you need to do the following
    1. c*v=n so 0.02*0.025= 0.0005 moles [this gives you the no. of moles in 25cm3]
    2. n/v=c so (0.0005)/0.1= 0.005 [this gives you the 'new' conc. of HCl in 100cm3 ]
    3. now simply -log(0.005)= 2.301
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    (Original post by freetown)
    You can't -log(0.02) because the HCl concentration is more dilute when you add the water.

    Because of this you need to do the following
    1. c*v=n so 0.02*0.025= 0.0005 moles [this gives you the no. of moles in 25cm3]
    2. n/v=c so (0.0005)/0.1= 0.005 [this gives you the 'new' conc. of HCl in 100cm3 ]
    3. now simply -log(0.005)= 2.301
    ahhhh thanks for your comprehensive reply!
 
 
 
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