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Maurice
Maurice
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As the amounts of substance in the final solution are known to be $n(\ce{NH3}) = \pu{2 mol},$$n(\ce{NH3}) = \pu{2 mmol},$ and $n(\ce{NH4^+}) = \pu{0.5 mol},$$n(\ce{NH4^+}) = \pu{0.5 mmol},$ you may simply use the definition of the constant $K_\mathrm{b}:$

$$K_\mathrm{b} = \frac{n(\ce{NH4^+})[\ce{OH^-}]}{n(\ce{NH3})} = \frac{\pu{0.5 mol}\times [\ce{OH-}]}{\pu{2 mol}} = \pu{3.3E-5}$$$$K_\mathrm{b} = \frac{n(\ce{NH4^+})[\ce{OH^-}]}{n(\ce{NH3})} = \frac{\pu{0.5 mmol}\times [\ce{OH-}]}{\pu{2 mmol}} = \pu{3.3E-5}$$

from where $[\ce{OH-}],$ $[\ce{H+}]$ and $\mathrm{pH}$ can be quickly obtained:

$$[\ce{OH-}] = \pu{1.32E-4 mol L^-1}$$

$$[\ce{H+}] = \frac{10^{-14}}{[\ce{OH-}]} = \pu{7.57E-11 mol L^-1}$$

$$\mathrm{pH} = -\log[\ce{H+}] = 10.12$$

As the amounts of substance in the final solution are known to be $n(\ce{NH3}) = \pu{2 mol},$ and $n(\ce{NH4^+}) = \pu{0.5 mol},$ you may simply use the definition of the constant $K_\mathrm{b}:$

$$K_\mathrm{b} = \frac{n(\ce{NH4^+})[\ce{OH^-}]}{n(\ce{NH3})} = \frac{\pu{0.5 mol}\times [\ce{OH-}]}{\pu{2 mol}} = \pu{3.3E-5}$$

from where $[\ce{OH-}],$ $[\ce{H+}]$ and $\mathrm{pH}$ can be quickly obtained:

$$[\ce{OH-}] = \pu{1.32E-4 mol L^-1}$$

$$[\ce{H+}] = \frac{10^{-14}}{[\ce{OH-}]} = \pu{7.57E-11 mol L^-1}$$

$$\mathrm{pH} = -\log[\ce{H+}] = 10.12$$

As the amounts of substance in the final solution are known to be $n(\ce{NH3}) = \pu{2 mmol},$ and $n(\ce{NH4^+}) = \pu{0.5 mmol},$ you may simply use the definition of the constant $K_\mathrm{b}:$

$$K_\mathrm{b} = \frac{n(\ce{NH4^+})[\ce{OH^-}]}{n(\ce{NH3})} = \frac{\pu{0.5 mmol}\times [\ce{OH-}]}{\pu{2 mmol}} = \pu{3.3E-5}$$

from where $[\ce{OH-}],$ $[\ce{H+}]$ and $\mathrm{pH}$ can be quickly obtained:

$$[\ce{OH-}] = \pu{1.32E-4 mol L^-1}$$

$$[\ce{H+}] = \frac{10^{-14}}{[\ce{OH-}]} = \pu{7.57E-11 mol L^-1}$$

$$\mathrm{pH} = -\log[\ce{H+}] = 10.12$$

Don't use an illiterate term "number of moles"; use \ce strictly for chemical formulae and not for the entire math expressions; don't ever omit units
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andselisk ♦
andselisk ♦
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As the numberamounts of molessubstance in the final solution are known to be $n \ce{(NH3) = 2}$ mol,$n(\ce{NH3}) = \pu{2 mol},$ and $n\ce{(NH4^+) = 0.5}$ mol,$n(\ce{NH4^+}) = \pu{0.5 mol},$ you may simply use the definition of the constant $K_b$ which is : $$\ce{K_b = \frac{n(NH4^+)[OH^-]}{n(NH3)} = \frac{0.5·[OH-]}{2} = 3.3·10^{-5}}$$$K_\mathrm{b}:$

$$K_\mathrm{b} = \frac{n(\ce{NH4^+})[\ce{OH^-}]}{n(\ce{NH3})} = \frac{\pu{0.5 mol}\times [\ce{OH-}]}{\pu{2 mol}} = \pu{3.3E-5}$$

from where $\ce{[OH-]}$,$[\ce{OH-}],$ $\ce{[H+]}$$[\ce{H+}]$ and $p$H$\mathrm{pH}$ can be quickly obtained  :$$\ce{[OH^-] = 1.32·10^{-4}}$$ $$\ce{[H^+] = \frac{10^{-14}}{[OH-]} = 7.57·10^{-11}}$$ $$p\ce{H = - log [H+] = 10.12}$$

$$[\ce{OH-}] = \pu{1.32E-4 mol L^-1}$$

$$[\ce{H+}] = \frac{10^{-14}}{[\ce{OH-}]} = \pu{7.57E-11 mol L^-1}$$

$$\mathrm{pH} = -\log[\ce{H+}] = 10.12$$

As the number of moles in the final solution are known to be $n \ce{(NH3) = 2}$ mol, and $n\ce{(NH4^+) = 0.5}$ mol, you may simply use the definition of the constant $K_b$ which is : $$\ce{K_b = \frac{n(NH4^+)[OH^-]}{n(NH3)} = \frac{0.5·[OH-]}{2} = 3.3·10^{-5}}$$ from where $\ce{[OH-]}$, $\ce{[H+]}$ and $p$H can be quickly obtained  :$$\ce{[OH^-] = 1.32·10^{-4}}$$ $$\ce{[H^+] = \frac{10^{-14}}{[OH-]} = 7.57·10^{-11}}$$ $$p\ce{H = - log [H+] = 10.12}$$

As the amounts of substance in the final solution are known to be $n(\ce{NH3}) = \pu{2 mol},$ and $n(\ce{NH4^+}) = \pu{0.5 mol},$ you may simply use the definition of the constant $K_\mathrm{b}:$

$$K_\mathrm{b} = \frac{n(\ce{NH4^+})[\ce{OH^-}]}{n(\ce{NH3})} = \frac{\pu{0.5 mol}\times [\ce{OH-}]}{\pu{2 mol}} = \pu{3.3E-5}$$

from where $[\ce{OH-}],$ $[\ce{H+}]$ and $\mathrm{pH}$ can be quickly obtained:

$$[\ce{OH-}] = \pu{1.32E-4 mol L^-1}$$

$$[\ce{H+}] = \frac{10^{-14}}{[\ce{OH-}]} = \pu{7.57E-11 mol L^-1}$$

$$\mathrm{pH} = -\log[\ce{H+}] = 10.12$$

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Maurice
Maurice
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As the number of moles in the final solution are known to be $n \ce{(NH3) = 2}$ mol, and $n\ce{(NH4^+) = 0.5}$ mol, you may simply use the definition of the constant $K_b$ which is : $$\ce{K_b = \frac{n(NH4^+)[OH^-]}{n(NH3)} = \frac{0.5·[OH-]}{2} = 3.3·10^{-5}}$$ from where $\ce{[OH-]}$, $\ce{[H+]}$ and $p$H can be quickly obtained :$$\ce{[OH^-] = 1.32·10^{-4}}$$ $$\ce{[H^+] = \frac{10^{-14}}{[OH-]} = 7.57·10^{-11}}$$ $$p\ce{H = - log [H+] = 10.12}$$