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First of all we have to calculate the pH of the solutionbefore addit.pdf

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First of all we have to calculate the pH of the solutionbefore addition of HNO3 According to Henderson\'s Equation , pOH = pKb + log ( [salt] / [ base] ) = - log Kb +log ( [salt] / [ base] ) = - log (1.8*10^-5 ) + log ( 0.52 / 0.10 ) = 5 - log 1.8 +0.716 = 5.46 So pH = 14 - pOH =8.54 Effect on pH when HNO3 is added : ------------------------------------- NH3 + HNO3 ---> NH4NO3 [ NH4NO3 ] = 0.1(from initial NH4NO3) + 0.01( fromaddition of HNO3) = 0.11 mol [NH3 ] = 0.1 - 0.01 = 0.09 moles pH = 14 - pOH = 14 -( pKb + log ( [salt] / [ base] ) ) = 14 -( 5- log 1.8 + log ( 0.11 / 0.09 ) =9.17 Solution First of all we have to calculate the pH of the solutionbefore addition of HNO3 According to Henderson\'s Equation , pOH = pKb + log ( [salt] / [ base] ) = - log Kb +log ( [salt] / [ base] ) = - log (1.8*10^-5 ) + log ( 0.52 / 0.10 ) = 5 - log 1.8 +0.716 = 5.46 So pH = 14 - pOH =8.54 Effect on pH when HNO3 is added : ------------------------------------- NH3 + HNO3 ---> NH4NO3 [ NH4NO3 ] = 0.1(from initial NH4NO3) + 0.01( fromaddition of HNO3) = 0.11 mol [NH3 ] = 0.1 - 0.01 = 0.09 moles pH = 14 - pOH = 14 -( pKb + log ( [salt] / [ base] ) ) = 14 -( 5- log 1.8 + log ( 0.11 / 0.09 ) =9.17.

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First of all we have to calculate the pH of the solutionbefore addition of HNO3 According to Henderson's Equation , pOH = pKb + log ( [salt] / [ base] ) = - log Kb +log ( [salt] / [ base] ) = - log (1.8*10^-5 ) + log ( 0.52 / 0.10 ) = 5 - log 1.8 +0.716 = 5.46 So pH = 14 - pOH =8.54 Effect on pH when HNO3 is added : ------------------------------------- NH3 + HNO3 ---> NH4NO3 [ NH4NO3 ] = 0.1(from initial NH4NO3) + 0.01( fromaddition of HNO3) = 0.11 mol [NH3 ] = 0.1 - 0.01 = 0.09 moles pH = 14 - pOH = 14 -( pKb + log ( [salt] / [ base] ) ) = 14 -( 5- log 1.8 + log ( 0.11 / 0.09 ) =9.17 Solution First of all we have to calculate the pH of the solutionbefore addition of HNO3 According to Henderson's Equation , pOH = pKb + log ( [salt] / [ base] ) = - log Kb +log ( [salt] / [ base] ) = - log (1.8*10^-5 ) + log ( 0.52 / 0.10 ) = 5 - log 1.8 +0.716 = 5.46 So pH = 14 - pOH =8.54 Effect on pH when HNO3 is added : ------------------------------------- NH3 + HNO3 ---> NH4NO3 [ NH4NO3 ] = 0.1(from initial NH4NO3) + 0.01( fromaddition of HNO3) = 0.11 mol [NH3 ] = 0.1 - 0.01 = 0.09 moles
pH = 14 - pOH = 14 -( pKb + log ( [salt] / [ base] ) ) = 14 -( 5- log 1.8 + log ( 0.11 / 0.09 ) =9.17

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First of all we have to calculate the pH of the solutionbefore addit.pdf

  • 1. First of all we have to calculate the pH of the solutionbefore addition of HNO3 According to Henderson's Equation , pOH = pKb + log ( [salt] / [ base] ) = - log Kb +log ( [salt] / [ base] ) = - log (1.8*10^-5 ) + log ( 0.52 / 0.10 ) = 5 - log 1.8 +0.716 = 5.46 So pH = 14 - pOH =8.54 Effect on pH when HNO3 is added : ------------------------------------- NH3 + HNO3 ---> NH4NO3 [ NH4NO3 ] = 0.1(from initial NH4NO3) + 0.01( fromaddition of HNO3) = 0.11 mol [NH3 ] = 0.1 - 0.01 = 0.09 moles pH = 14 - pOH = 14 -( pKb + log ( [salt] / [ base] ) ) = 14 -( 5- log 1.8 + log ( 0.11 / 0.09 ) =9.17 Solution First of all we have to calculate the pH of the solutionbefore addition of HNO3 According to Henderson's Equation , pOH = pKb + log ( [salt] / [ base] ) = - log Kb +log ( [salt] / [ base] ) = - log (1.8*10^-5 ) + log ( 0.52 / 0.10 ) = 5 - log 1.8 +0.716 = 5.46 So pH = 14 - pOH =8.54 Effect on pH when HNO3 is added : ------------------------------------- NH3 + HNO3 ---> NH4NO3 [ NH4NO3 ] = 0.1(from initial NH4NO3) + 0.01( fromaddition of HNO3) = 0.11 mol [NH3 ] = 0.1 - 0.01 = 0.09 moles
  • 2. pH = 14 - pOH = 14 -( pKb + log ( [salt] / [ base] ) ) = 14 -( 5- log 1.8 + log ( 0.11 / 0.09 ) =9.17