Em tham khảo nha:

\(\begin{array}{l}
a)\\
{n_{{K_2}S{O_4}}} = \dfrac{{17,4}}{{174}} = 0,1\,mol\\
{n_{{K^ + }}} = 2{n_{{K_2}S{O_4}}} = 0,2\,mol\\
{n_{S{O_4}^{2 - }}} = {n_{{K_2}S{O_4}}} = 0,1\,mol\\
{\rm{[}}{K^ + }{\rm{]}} = \dfrac{{0,2}}{2} = 0,1M\\
{\rm{[}}S{O_4}^{2 - }{\rm{]}} = \dfrac{{0,1}}{2} = 0,05M\\
b)\\
{n_{NaOH}} = 0,25 \times 0,2 = 0,05\,mol\\
{n_{NaCl}} = 0,25 \times 0,1 = 0,025\,mol\\
{n_{N{a^ + }}} = 0,05 + 0,025 = 0,075\,mol\\
{n_{O{H^ - }}} = {n_{NaOH}} = 0,05\,mol\\
{n_{C{l^ - }}} = {n_{NaCl}} = 0,025\,mol\\
{V_{{\rm{dd}}}} = 0,25 + 0,25 = 0,5l\\
{\rm{[}}N{a^ + }{\rm{]}} = \dfrac{{0,075}}{{0,5}} = 0,15M\\
{\rm{[}}O{H^ - }{\rm{]}} = \dfrac{{0,05}}{{0,5}} = 0,1M\\
{\rm{[}}C{l^ - }{\rm{]}} = \dfrac{{0,025}}{{0,5}} = 0,05M\\
c)\\
{\rm{[}}{H^ + }{\rm{]}} = {10^{ - 4}} = 0,0001M\\
{\rm{[}}S{O_4}^{2 - }{\rm{]}} = \dfrac{{0,0001}}{2} = 0,00005M\\
d)\\
{\rm{[}}O{H^ - }{\rm{]}} = {10^{ - (14 - 11)}} = {10^{ - 3}} = 0,001M\\
{\rm{[}}{K^ + }{\rm{]}} = {\rm{[}}O{H^ - }{\rm{]}} = 0,001M
\end{array}\)

Đáp án:

\(\begin{array}{l} a,\ [K^+]=0,1\ (M).\\ [SO_4^{2-}]=0,05\ (M).\\ b,\ [Na^+]=0,15\ (M).\\ [Cl^-]=0,05\ (M).\\ [OH^-]=0,1\ (M).\\ c,\ [H^+]=0,0001\ (M).\\ [SO_4^{2-}]=0,00005\ (M).\\ d,\ [OH^-]=0,001\ (M).\\ [K^+]=0,001\ (M).\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l} a,\\ \text{PT ion:}\ K_2SO_4\to 2K^++SO_4^{2-}\\ n_{K_2SO_4}=\dfrac{17,4}{174}=0,1\ (mol).\\ Theo\ pt:\\ n_{K^+}=2n_{K_2SO_4}=0,2\ (mol).\\ n_{SO_4^{2-}}=n_{K_2SO_4}=0,1\ (mol).\\ \Rightarrow [K^+]=\dfrac{0,2}{2}=0,1\ (M).\\ [SO_4^{2-}]=\dfrac{0,1}{2}=0,05\ (M).\\ b,\\ \text{PT ion:}\\ NaCl\to Na^++Cl^-\ (1)\\ NaOH\to Na^++OH^-\ (2)\\ n_{NaCl}=0,25\times 0,1=0,025\ (mol).\\ n_{NaOH}=0,25\times 0,2=0,05\ (mol).\\ Theo\ pt\ (1),\ (2):\\ n_{Na^{+}}=n_{NaCl}+n_{NaOH}=0,025+0,05=0,075\ (mol).\\ n_{Cl^-}=n_{NaCl}=0,025\ (mol).\\ n_{OH^-}=n_{NaOH}=0,05\ (mol).\\ V_{\text{dd}}=V_{\text{dd NaCl}}+V_{\text{dd NaOH}}=250+250=500\ (ml)=0,5\ (l).\\ \Rightarrow [Na^+]=\dfrac{0,075}{0,5}=0,15\ (M).\\ [Cl^-]=\dfrac{0,025}{0,5}=0,05\ (M).\\ [OH^-]=\dfrac{0,05}{0,5}=0,1\ (M).\\ c,\\ \text{PT ion:}\ H_2SO_4\to 2H^++SO_4^{2-}\\ pH=4\ \Rightarrow [H^+]=10^{-4}=0,0001\ (M).\\ \Rightarrow [SO_4^{2-}]=\dfrac{1}{2}[H^+]=\dfrac{0,0001}{2}=0,00005\ (M).\\ d,\\ \text{PT ion:}\ KOH\to K^++OH^-\\ pH=11\ \Rightarrow pOH=14-pH=14-11=3\\ \Rightarrow [OH^-]=10^{-3}=0,001\ (M).\\ \Rightarrow [K^+]=[OH^-]=0,001\ (M).\end{array}\)

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