Tính pH của dung dịch A, biết KNH4+=10^-9.24

1. a)

n_NH3=0,03(mol) n_HCl=0,015(mol)

\(\begin{matrix}NH_3&+HCl&\rightarrow&NH_4Cl&\\0,015&0,015&&0,015&\left(mol\right)\end{matrix}\)

\(\begin{matrix}NH_4Cl&\rightarrow&NH_4^+&+Cl^-&\\0,015&&0,15&0,015&\left(mol\right)\end{matrix}\)

Gọi x là số mol NH₃ phân li(x>0), k_a là hằng số axit của NH₄⁺, k_b là hằng số bazo của NH₃. Theo đề ta có:

V_dd=0,05+0,15=0,2(l)

\(NH_4^+\rightarrow NH_3+H^+\left(k_a=10^{-9,24}\right)\)

⇒\(\begin{matrix}&NH_3&+H_2O\rightarrow&NH_4^+&+OH^-&k_b=\dfrac{10^{-14}}{k_a}=\dfrac{10^{-14}}{10^{-9,24}}=10^{-4,76}\\BD&0,015&&0,015&&\\PU&x&&x&x&\left(mol\right)\\CB&0,015-x&&0,015+x&x&\end{matrix}\begin{matrix}\\\\\\\end{matrix}\)

Ta có: \(k_B=\dfrac{\left[NH_4^+\right]\left[OH^-\right]}{\left[NH_3\right]}=\dfrac{\left(\dfrac{0,015+x}{0,2}\right)\left(\dfrac{x}{0,2}\right)}{\dfrac{0,015-x}{0,2}}=10^{-4,76}\Rightarrow x=3,47399.10^{-6}\left(x>0\right)\)

⇒\(\left[OH^-\right]=\dfrac{3,47399.10^{-6}}{0,2}=1,736995.10^{-5}M\)

⇒\(pOH=-\log\left(\left[OH^-\right]\right)=-\log\left(1,736995.10^{-5}\right)=4,76020\)

⇒\(pH=14-pOH=14-4,76020=9,2398\)

b)

Thêm vào dung dịch 1,5.10^-4 (mol) HCl:

NH₃ dư tiếp tục tác dụng:

\(\begin{matrix}NH_3&+HCl&\rightarrow&NH_4Cl&&\\1,5.10^{-4}&1,5.10^{-4}&&1,5.10^{-4}&\left(mol\right)&\end{matrix}\)

\(\begin{matrix}NH_4Cl&\rightarrow&NH^+_4&+Cl^-&&\\1,5.10^{-4}&&1,5.10^{-4}&1,5.10^{-4}&\left(mol\right)&\end{matrix}\)

Gọi y là số mol NH₃ phân li ở trường hợp b (y>0)

\(\begin{matrix}&NH_3&+H_2O\rightarrow&NH_4^+&+OH^-&\\BD&0,01485&&0,01515&&\\PU&y&&y&y&\left(mol\right)\\CB&0,01485-y&&0,01515+y&y&\end{matrix}\begin{matrix}k_b=10^{-4,76}\\\\\\\end{matrix}\)

Ta có: \(k_B=\dfrac{\left[NH_4^+\right]\left[OH^-\right]}{\left[NH_3\right]}=\dfrac{\left(\dfrac{0,01515+y}{0,2}\right)\left(\dfrac{y}{0,2}\right)}{\dfrac{0,01485-y}{0,2}}=10^{-4,76}\Rightarrow y=3,40523.10^{-6}\left(y>0\right)\)

⇒\(\left[OH^-\right]=\dfrac{3,40523.10^{-6}}{0,2}=1,70262.10^{-5}M\)

⇒\(pOH=-\log\left(\left[OH^-\right]\right)=-\log\left(1,70262.10^{-5}\right)=4,76888\)

⇒\(pH=14-pOH=14-4,76888=9,23112\)

c)

Thêm 1,5.10^-4 mol NaOH

\(\begin{matrix}NH_4Cl&+NaOH&\rightarrow&NaCl&+NH_3&+H_2O\\1,5.10^{-4}&1,5.10^{-4}&&&1,5.10^{-4}&\end{matrix}\left(mol\right)\)

Gọi z là số mol NH₃ phân li trong trường hợp c (z>0)

\(\begin{matrix}&NH_3&+H_2O\rightarrow&NH_4^+&+OH^-&\\BD&0,01515&&0,01485&&\\PU&z&&z&z&\left(mol\right)\\CB&0,01515-z&&0,01485+z&z&\end{matrix}\begin{matrix}k_b=10^{-4,76}\\\\\\\end{matrix}\)

Ta có: \(k_B=\dfrac{\left[NH_4^+\right]\left[OH^-\right]}{\left[NH_3\right]}=\dfrac{\left(\dfrac{0,01485+z}{0,2}\right)\left(\dfrac{z}{0,2}\right)}{\dfrac{0,01515-z}{0,2}}=10^{-4,76}\Rightarrow z=3,54414.10^{-6}\left(z>0\right)\)

⇒\(\left[OH^-\right]=\dfrac{3,54414.10^{-6}}{0,2}=1,77207.10^{-5}M\)

⇒\(pOH=-\log\left(\left[OH^-\right]\right)=-\log\left(1,77207.10^{-5}\right)=4,75152\)

⇒\(pH=14-pOH=14-4,75152=9,24848\)