\(n_{NaOH}=0.015\cdot2=0.03\left(mol\right)\)

\(n_{H_2SO_4}=0.015\cdot1.5=0.0225\left(mol\right)\)

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\)

\(0.03..........0.015...........0.015\)

\(n_{H_2SO_4\left(dư\right)}=0.0225-0.015=0.0075\left(mol\right)\)

\(C_{M_{Na^+}}=\dfrac{0.015\cdot2}{0.015+0.015}=1\left(M\right)\)

\(C_{M_{H^+}}=\dfrac{0.0075\cdot2}{0.015+0.015}=0.5\left(M\right)\)

\(C_{M_{SO_4^{2-}}}=\dfrac{0.015+0.0075}{0.015+0.015}=0.75\left(M\right)\)

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 \(0,0075\left(mol\right)\)

\(\Rightarrow\dfrac{H_2SO_4}{0,0075}\dfrac{\rightarrow}{ }\dfrac{2H^+}{0,015}\dfrac{+}{ }\dfrac{SO_4^{2-}}{ }\) \(\Rightarrow\left[H^+\right]=\dfrac{0,015}{0,03}=0,5\)

vậy .................................................................................................

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 .................................................................................................

\(a.n_{HCl}=0,1.1=0,1\left(mol\right)\\ n_{H_2SO_4}=0,5.0,1=0,05\left(mol\right)\\ \left[HCl\right]=\dfrac{0,1}{0,1+0,1}=0,5\left(M\right)\\ \left[H_2SO_4\right]=\dfrac{0,05}{0,1+0,1}=0,25\left(M\right)\\ \left[H^+\right]=0,5+0,25.2=1\left(M\right)\\ \left[SO^{2-}_4\right]=\left[H_2SO_4\right]=0,25\left(M\right)\\ \left[Cl^-\right]=\left[HCl\right]=0,5\left(M\right)\)

\(b.BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\\ n_{BaSO_4}=n_{H_2SO_4}=0,05\left(mol\right)\\ m_{\downarrow}=m_{BaSO_4}=233.0,05=11,65\left(g\right)\)

\(a.n_{NaOH}=1.0,15=0,15\left(mol\right)\\ n_{KOH}=0,5.0,1=0,05\left(mol\right)\\ \left[Na^+\right]=\left[NaOH\right]=\dfrac{0,15}{0,15+0,1}=0,6\left(M\right)\\ \left[K^+\right]=\left[KOH\right]=\dfrac{0,05}{0,1+0,15}=0,2\left(M\right)\\ \left[OH^-\right]=0,2+0,6=0,8\left(M\right)\\ b.2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\\ 2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ n_{H_2SO_4}=\dfrac{1}{2}.\left(n_{KOH}+n_{NaOH}\right)=\dfrac{0,15+0,05}{2}=0,1\left(mol\right)\\ a=C_{MddH_2SO_4}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)

\(n_{NaOH}=0,015.2=0,03\left(mol\right)\)

\(n_{H_2SO_4}=0,01.1,5=0,015\left(mol\right)\)

PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

Xét: \(\frac{n_{NaOH}}{2}=0,015=\frac{n_{H_2SO_4}}{1}=0,015\)\

=> NaOH và H₂SO₄ vừa đủ

=> Dung dịch sau phản ứng trung hòa

Tính C_M của chất nào vậy

\(n_{BaSO_4}=\frac{m}{M}=\frac{32,62}{233}=0,14mol\)

PTHH:

\(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4+2H_2O\)

0,14 0,14 0,14 0,28 (mol)

Gọi \(V_{ddH_2SO_4}\)cần thêm là x

\(n_{H_2SO_4}=\frac{m}{M}=\frac{98}{98}=1mol\)

\(C^{\left(A\right)}_{M_{H_2SO_4}}=\frac{1}{1}=1M\)

\(n^{\left(A\right)}_{H2SO4}=C_M.V=1.x=xmol\)

\(n_{H2SO4}=C_M.V=2.0,4=0,8mol\)

\(C_{MX}=\frac{n}{V}=\frac{0,8+x}{0,4+x}\left(M\right)\)

\(n_X=C_{MX}.V\)

\(\Leftrightarrow0,14=\frac{0,8+x}{0,4+x}.0,1\)

\(\Leftrightarrow\frac{0,14}{0,1}=\frac{0,8+x}{0,4+x}\)

⇔0,08+0,1x=0,56+0,14x

⇔x=0,6(l)

Vậy cần thêm 0,6 l dung dịch

\(a.n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\\ \left[Na^+\right]=\left[OH^-\right]=\left[NaOH\right]=\dfrac{0,5}{0,5}=1\left(M\right)\\ b.HCl+NaOH\rightarrow NaCl+H_2O\\ n_{HCl}=n_{NaOH}=0,5\left(mol\right)\\ V_{ddHCl}=\dfrac{0,5}{2}=0,25\left(l\right)\)