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\(n_{H_2SO_4}=0,05.0,4=0,02\left(mol\right)\\ n_{HCl}=0,35.0,2=0,07\left(mol\right)\\ \left[H_2SO_4\right]=\dfrac{0,02}{0,05+0,35}=0,05\left(M\right)\\ \left[HCl\right]=\dfrac{0,07}{0,05+0,35}=0,175\left(M\right)\\ \Rightarrow\left[H^+\right]=0,05.2+0,175.1=0,275\left(M\right)\\ \left[SO^{2-}_4\right]=0,05\left(M\right)\\ \left[Cl^-\right]=0,175\left(M\right)\)
\(n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\\ n_{H_2SO_4}=\dfrac{9,8}{98}=0,1\left(mol\right)\\ \left[HCl\right]=\dfrac{0,2}{0,2}=1\left(M\right)\\ \left[H_2SO_4\right]=\dfrac{0,1}{0,2}=0,5\left(M\right)\\ \left[SO_4^{2-}\right]=\left[H_2SO_4\right]=0,5\left(M\right)\\ \left[Cl^-\right]=\left[HCl\right]=1\left(M\right)\\ \left[H^+\right]=1+0,5.2=2\left(M\right)\)
PT ion: \(H^++OH^-\rightarrow H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{H^+}=0,2\cdot0,5\cdot2=0,2\left(mol\right)\\n_{OH^-}=0,05\cdot2=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) H+ còn dư 0,1 mol
\(\Rightarrow\left[H^+\right]=\dfrac{0,1}{0,25}=0,4\left(M\right)\) \(\Rightarrow pH=-log\left(0,4\right)\approx0,4\)
câu 1 : ta có : \(\dfrac{2NaOH}{0,03}\dfrac{+}{ }\dfrac{H_2SO_4}{0,0225}\dfrac{\rightarrow}{ }\dfrac{Na_2SO_4}{ }\dfrac{+}{ }\dfrac{2H_2O}{ }\)
\(\Rightarrow NaOH\) phản ứng hết và \(H_2SO_4\) dư \(0,0075\left(mol\right)\)
\(\Rightarrow\dfrac{H_2SO_4}{0,0075}\dfrac{\rightarrow}{ }\dfrac{2H^+}{0,015}\dfrac{+}{ }\dfrac{SO_4^{2-}}{ }\) \(\Rightarrow\left[H^+\right]=\dfrac{0,015}{0,03}=0,5\)
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câu 2 : ta có : \(\dfrac{2KOH}{0,1}\dfrac{+}{ }\dfrac{H_2SO_4}{0,05}\dfrac{\rightarrow}{ }\dfrac{k_2SO_4}{0,05}\dfrac{+}{ }\dfrac{2H_2O}{0,1}\)
\(\Rightarrow m_{chấtrắng}=m_{K_2SO_4}+m_{KOH_{dư}}\) \(\Leftrightarrow m_{KOH_{dư}}=m_{chấtrắng}-m_{K_2SO_4}\)
\(\Leftrightarrow m_{KOH_{dư}}=11,5-0,05.174=2,8\)
\(\Rightarrow m_{KOH}=0,1.56+2,8=3,36\) \(\Rightarrow n_{KOH}=\dfrac{3,36}{56}=0,06\)
\(\Rightarrow C_M=\dfrac{0,06}{0,15}=0,4\left(M\right)\)
vậy .................................................................................................
\(n_{NaOH}=0,02.2=0,04\left(mol\right)\\ 2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ a.n_{H_2SO_4}=n_{Na_2SO_4}=\dfrac{0,04}{2}=0,02\left(mol\right)\\ C_{MddH_2SO_4}=\dfrac{0,02}{0,08}=0,25\left(M\right)\\ b.\left[Na^+\right]=\dfrac{0,02.2}{0,02+0,08}=0,4\left(M\right)\\ \left[SO^{2-}_4\right]=\dfrac{0,02}{0,02+0,08}=0,2\left(M\right)\)
\(n_{H_2SO_4\left(2M\right)}=0,15.2=0,3\left(mol\right)\)
\(n_{H_2SO_4\left(3M\right)}=0,15.3=0,45\left(mol\right)\)
\(n_{H_2SO_4\left(B\right)}=0,3+0,45=0,75\left(mol\right)\)
\(\Rightarrow C_{M_{ddB}}=\dfrac{0,75}{0,2}=3,75M\)
\(n_{H_2SO_4\left(tổng\right)}=0,15.2+0,05.3=0,45\left(mol\right)\\ V_{ddH_2SO_4\left(tổng\right)}=150+50=200\left(ml\right)=0,2\left(l\right)\\ C_{MddH_2SO_4\left(sau\right)}=C_{MddB}=\dfrac{0,45}{0,2}=2,25\left(M\right)\)