5.1.3~ acids, bases and buffers

Cards (9)

  • pH of a buffer: weak acid + strong base
    1. Work out the moles of acid and base
    2. identify if the base or acid is in excess
    3. limiting reagent tells us the mole of the salt [A^-]
    4. if the base is in excess work out the moles unreacted and the new concentration [OHOH^-]=moles unreactednew volume \frac{moles\ unreacted}{new\ volume\ }
    5. [H+]\left[H^+\right]= Kw[OH]\frac{Kw}{\left[OH^-\right]}
  • What do you then do if the acid is in excess
    [HA]=moles of HA unreacted new volume \frac{moles\ of\ HA\ unreacted\ }{new\ volume\ }
    [A^-]=moles of salt formed [A]new volume\frac{moles\ of\ salt\ formed\ \left[A^-\right]}{new\ volume}
    then [H+H^+]=Ka[HA][A]\frac{Ka\left[HA\right]}{\left[A^-\right]}
    then -log[H+H^+]
  • pH of a buffer: strong acid + strong base
    1. work out the moles of the acid and the base
    2. identify if it is the acid or base that is in excess and work out the left over moles/moles unreacted
    3. if the base is in excess work out the new conc
    4. new concentration=moles unreactedtotal volume \frac{moles\ unreacted}{total\ volume\ }
    5. then [H+]\left[H^+\right]=Kw[OH]\frac{Kw}{\left[OH^-\right]}
  • What to you do when if the acid is in excess
    new concentration=moles unreactedtotal volume\frac{moles\ unreacted}{total\ volume}
    then -log[H+]\left[H^+\right]
  • calculating the pH of water
    in pure water [H+]=\left[H^+\right]=[OH]\left[OH^-\right]
    therefore Kw=[H+]2\left[H^+\right]^2
    therefore [H+]\left[H^+\right]= Kw\sqrt{Kw}
  • What is the effect of temperature on the pH of water and the neutrality of water
    as the temperature increase the equilibrium moves to the right to lower the temperature
    therefore [H+]and[OH]\left[H^+\right]and\left[OH^-\right] increase
    therefore Kw increases and the pH also increase however the water is still neutral as [H+]=\left[H^+\right]=[OH]\left[OH^-\right]
  • How do you calculate the pH of a strong base
    find [OH]\left[OH^-\right] concentration
    then substitute into equation [H+]\left[H^+\right]=Kw[OH]\frac{Kw}{\left[OH^-\right]}
    whereby the value of Kw at room temperature is 1×10141\times10^{-14}
  • How do you calculate the pH of a diluted strong acid
    [H+]\left[H^+\right]=[H+]old\left[H^+\right]_{old}×\timesold volumenew volume\frac{old\ volume}{new\ volume}
  • What is the relationship between the strength of an acid and its Ka and pKa value
    the bigger the value of Ka the stronger the acid
    the bigger the value of pKa the weaker the acid