Đáp án A

\(n_{AgNO_3}=0,1.0,2=0,02\left(mol\right)\\ n_{HCl}=0,3.0,2=0,06\left(mol\right)\\ AgNO_3+HCl\rightarrow AgCl\downarrow\left(trắng\right)+HNO_3\\ a,Vì:\dfrac{0,02}{1}< \dfrac{0,06}{1}\Rightarrow HCldư\\ \Rightarrow n_{AgCl}=n_{HNO_3}=n_{AgNO_3}=0,02\left(mol\right)\\ \Rightarrow m_{\downarrow}=m_{AgCl}=143,5.0,02=2,87\left(g\right)\\ b,dd.sau.p.ứ:HNO_3,HCl\left(dư\right)\\ n_{HCl\left(dư\right)}=0,06-0,02=0,04\left(mol\right)\\V_{ddsau}=V_{ddAgNO_3}+V_{ddHCl}=0,2+0,2=0,4\left(l\right)\\ \Rightarrow C_{MddHCl\left(dư\right)}=\dfrac{0,04}{0,4}=0,1\left(M\right)\\ C_{MddHNO_3}=\dfrac{0,02}{0,4}=0,05\left(M\right)\)

a)
$BaCl_2 + Na_2SO_4 \to BaSO_4 + 2NaCl$

$n_{BaCl_2} = 0,01 = n_{Na_2SO_4} = 0,01 \Rightarrow $ Vừa đủ

$n_{BaSO_4} = n_{Na_2SO_4} = 0,01(mol)$
$m_{BaSO_4} = 0,01.233 = 0,233(gam)$
b)

$n_{NaCl} = 2n_{Na_2SO_4} = 0,02(mol)$
$V_{dd} = 0,1 + 0,2 = 0,3(lít)$
$C_{M_{NaCl}} = \dfrac{0,02}{0,3} = 0,067M$

c)

$[Na^+] = [Cl^-] = C_{M_{NaCl}} = 0,067M$

\(n_{BaCl_2}=0.1\cdot0.1=0.01\left(mol\right)\)

\(n_{Na_2SO_4}=0.2\cdot0.05=0.01\left(mol\right)\)

\(BaCl_2+Na_2SO_4\rightarrow BaSO_4+2NaCl\)

\(0.01..........0.01............0.01..............0.02\)

\(m_{BaSO_4}=0.01\cdot233=2.33\left(g\right)\)

\(C_{M_{NaCl}}=\dfrac{0.01}{0.1+0.2}=0.03\left(M\right)\)

\(\left[Na^+\right]=\left[Cl^-\right]=0.03\left(M\right)\)

a, \(n_{HCl}=0,2.0,1=0,02\left(mol\right)=n_{H^+}=n_{Cl^-}\)

\(n_{H_2SO_4}=0,2.0,15=0,03\left(mol\right)=n_{SO_4^{2-}}\) \(\Rightarrow n_{H^+}=2n_{H_2SO_4}=0,06\left(mol\right)\)

\(\Rightarrow\Sigma n_{H^+}=0,02+0,06=0,08\left(mol\right)\)

\(n_{Ba\left(OH\right)_2}=0,3.0,05=0,015\left(mol\right)=n_{Ba^{2+}}\) 

\(\Rightarrow n_{OH^-}=2n_{Ba\left(OH\right)_2}=0,03\left(mol\right)\)

\(H^++OH^-\rightarrow H_2O\)

0,03___0,03 (mol) ⇒ n_H⁺ dư = 0,05 (mol)

\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\)

0,015___0,015______0,015 (mol) ⇒ n_SO4^2- dư = 0,015 (mol)

⇒ m = m_BaSO4 = 0,015.233 = 3,495 (g)

\(\left[Cl^-\right]=\dfrac{0,02}{0,2+0,3}=0,04\left(M\right)\)

\(\left[H^+\right]=\dfrac{0,05}{0,2+0,3}=0,1\left(M\right)\)

\(\left[SO_4^{2-}\right]=\dfrac{0,015}{0,2+0,3}=0,03\left(M\right)\)

b, pH = -log[H⁺] = 1

PTHH: \(Na_2SO_4+CaCl_2\rightarrow2NaCl+CaSO_4\downarrow\)

Ta có: \(\left\{{}\begin{matrix}n_{Na_2SO_4}=0,1\cdot0,5=0,05\left(mol\right)\\n_{CaCl_2}=0,1\cdot0,4=0,04\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Na₂SO₄ dư

\(\Rightarrow\left\{{}\begin{matrix}n_{CaSO_4}=0,04\left(mol\right)\\n_{NaCl}=0,08\left(mol\right)\\n_{Na_2SO_4\left(dư\right)}=0,01\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaSO_4}=0,04\cdot136=5,44\left(g\right)\\C_{M_{NaCl}}=\dfrac{0,08}{0,1+0,1}=0,4\left(M\right)\\C_{M_{Na_2SO_4\left(dư\right)}}=\dfrac{0,01}{0,2}=0,05\left(M\right)\end{matrix}\right.\)

a, \(n_{Ba\left(OH\right)_2}=0,1.0,1=0,01\left(mol\right)=n_{Ba^{2+}}\)

\(\Rightarrow n_{OH^-}=2n_{Ba\left(OH\right)_2}=0,02\left(mol\right)\)

\(n_{NaOH}=0,1.0,1=0,01\left(mol\right)=n_{Na^+}=n_{OH^-}\)

⇒ Σn_OH^- = 0,02 + 0,01 = 0,03 (mol)

\(n_{H_2SO_4}=0,4.0,0175=0,007\left(mol\right)=n_{SO_4^{2-}}\)

\(\Rightarrow n_{H^+}=2n_{H_2SO_4}=0,014\left(mol\right)\)

\(H^++OH^-\rightarrow H_2O\)

0,014___0,014 (mol) ⇒ n_OH^- dư = 0,03 - 0,014 = 0,016 (mol)

\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\)

0,007____0,007_____0,007 (mol) ⇒ n_Ba²⁺ dư = 0,01 - 0,007 = 0,003 (mol)

⇒ m = 0,007.233 = 1,631 (g)

\(\left[OH^-\right]=\dfrac{0,016}{0,1+0,4}=0,032\left(M\right)\)

\(\left[Ba^{2+}\right]=\dfrac{0,003}{0,1+0,4}=0,006\left(M\right)\)

\(\left[Na^+\right]=\dfrac{0,01}{0,1+0,4}=0,02\left(M\right)\)

b, pH = 14 - (-log[OH^-]) ≃ 12,505

\(n_{Ba^{2+}}=0,1.0,1=0,01\left(mol\right)\)

\(n_{SO_4^{2-}}=0,4.0,0175=7.10 ^{-3}\left(mol\right)\)

\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\downarrow\)

\(\Rightarrow m=m_{BaSO_4}=7.10^{-3}.233=1,631\left(g\right)\)

Ta có:

\(n_{H^+}=0,4.0,0175.2=0,014\left(mol\right)\)

\(n_{OH^-}=0,1.0,1.2+0,1.0,1=0,03\left(mol\right)\)

Trong dung dịch X:

\(n_{OH^-}=0,03-0,014=0,016\left(mol\right)\)\(\Rightarrow\left[OH^-\right]=\dfrac{0,016}{0,1+0,4}=0,032\left(M\right)\)

\(n_{Ba^{2+}}=0,01-7.10^{-3}=3.10^{-3}\left(mol\right)\Rightarrow\left[Ba^{2+}\right]=\dfrac{3.10^{-3}}{0,1+0,4}=6.10^{-3}\left(M\right)\)

\(n_{Na^+}=0,1.0,1=0,01\left(mol\right)\Rightarrow\left[Na^+\right]=0,02\)

\(pOH=-lg\left(0,032\right)\approx1,5\Rightarrow pH=14-1,5=12,5\)

nH⁺=0,04 mol      nOH^-=0,03 mol

H⁺              +        OH^-     -------->      H20

0,04                     0,03

0,03                     0,03                       0,03

0,01

a/ [H+] du=0,01/0,2=0,05 M

[SO4^2-]=0,01/0,2=0,05 M

[K⁺]=0,01/0,2=0,05 M

[Ba²⁺]=0,01/0,2=0,05M

b/ nH⁺ du=0,01/0,2=0,05 M

pH=-log(0,05)=1,3

c/ khoi luong chat ran thu duoc sau phan ung la

mcr= mSO4^2-  +  mK⁺    +  mBa²⁺

      =0,01.96+0,01.39+0,01.137

       =2,72g

ta có : \(\Sigma n_{H^+}=n_{HCl}+2n_{H_2SO_4}=0,04\left(mol\right)\)

\(\Sigma n_{OH^-}=n_{KOH}+2n_{Ba\left(OH\right)_2}=0,03\left(mol\right)\)

\(n_{SO_4^{2-}}=0,01\left(mol\right)\) ; \(n_{Ba^{2+}}=0,01\left(mol\right)\)

a, PT : \(H^++OH^-\rightarrow H_2O\)

0,03 0,03 0,03 (mol)

\(\Rightarrow n_{H^+}dư=0,01\left(mol\right)\)

đến đây tự tính đc nha. dùng ct \(CM=\dfrac{n}{V}\)

b, \(PH=-log[H^+]=-log\left(\dfrac{0,01}{0,2}\right)\simeq1,3\)

c, \(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\downarrow\)

0,01 0,01 0,01 (mol)\(mcr=m\downarrow+m_{K^+}=m_{BaSO_4}+m_{K+}=\left(0,01\times233\right)+\left(0,01\times39\right)=2,72\left(g\right)\)