Question

A 75.0-mLmL volume of 0.200 MM NH3NH3 (Kb=1.8×10−5Kb=1.8×10−5) is titrated with 0.500 MM HNO3HNO3. Calculate the pHpH after the addition of 13.0 mLmL of HNO3

Answer 1

The pH is
`pH = 9.4`

Step-by-step explanation:

From the question we are told that

The volume of
`NH_3` is
`V_N = 75mL = 75 *10^(-3) L`

The concentration of
`NH_3` is
`C_N = 0.200M`

The concentration of
`HNO_3` is
`C_H = 0.500 M`

The volume of
`HNO_3` added is
`V_H = 13mL = 13 *10^(-3 ) L`

The base dissociation constant is
`K_b = 1.8*10^(-5)`

The number of moles of
`HNO_3` that was titrated can be mathematically represented as

`n__(H)} = C_H * V_H`

substituting values

`n__(H)} = 0.500* 13*10^(-3)`

`n__(H)} = 0.0065 \ moles`

The number of moles of
`NH_3` that was titrated can be mathematically represented as

`n__(N)} = C_N * V_N`

substituting values

`n__(N)} = 0.200 * 75*10^(-3)`

`n__(N)} = 0.015 \ mole`

So from the calculation above the limited reactant is
`HNO_3`

The chemical equation for this reaction is

`NH_3 + HNO_3 ------> NH^(4+) + NO^(3+)`

From the chemical reaction

1 mole of
`HNO_3` is titrated with 1 mole of
`NH_3` to produce 1 mole of NH^{4+}

So

0.0065 moles of
`HNO_3` is titrated with 0.0065 mole of
`NH_3` to produce 0.0065 mole of
`NH^(4+)`

So

The remaining moles of
`NH_3` after the titration is

`n = n__(N)} - n__(H)}`

=>
`n = 0.015 - 0.0065`

`n = 0.0085 \ moles`

Now according to Henderson-Hasselbalch equation the pH of the reaction is mathematically represented as

`pH = pK_a + log [(NH_3)/(NH^(4+)) ]`

Where
`pK_b` is mathematically represented as

`pK_a = -log K_a`

Now
`K_a = (K_w)/(K_b)`

Where
`K_w` is the ionization constant of
`NH_3` with value
`K_w = 1.0*10^(-14)`

Hence
`K_a = (1.0*10^(-14))/(1.8 *10^(-5))`

`K_a = 5.556 * 10^(-10)`

Substituting this into the equation

`pH = -log K_a + log [(NH_3)/(NH^(4+)) ]`

`pH = log [((NH_3)/(NH^(4+)) )/(K_a) ]`

substituting values

`pH = log [((0.0085)/(0.0065) )/(5.556*10^(-10)) ]`

`pH = 9.4`