\(n_{HCl}=0,01.0,2=0,002\left(mol\right)\)

\(n_{HNO3}=0,03.0,3=0,009\left(mol\right)\)

\(\Rightarrow n_{H^+}=n_{HCl}+n_{HNO3}=0,002+0,009=0,011\left(mol\right)\)

\(\Rightarrow\left[H^+\right]=\dfrac{0,011}{0,2+0,3}=0,022M\Rightarrow pH=-log\left(0,022\right)=1,65\)

\(n_{NaOH}=0,1.0,01=0,001(mol)\\ \Rightarrow n_{OH^{-}}=0,001(mol)\\ n_{HCl}=0,03.0,2=0,006(mol)\\ \Rightarrow n_{H^{+}}=0,006(mol)\\ H^{+}+OH^{-}\to H_2O\\ 0,001<0,006\\ OH^{-} hêt; H^{+} dư\\ n_{H^{+}}=0,006-0,001=0,005(mol)\\ [H^{+}]=\frac{0,005}{0,1+0,2}=\frac{1}{60}M\\ \to pH=-log(\frac{1}{60})=1,77 \)

nHCl=0,001(mol)

nHNO3=0,003(mol)

=> nH⁺ =0,004(mol)

=>[H⁺] =\(\frac{0,004}{0,2}=0,02(M)\)

=> pH = -log(0,02)\(\approx\)1,7.

$n_{NaOH} = 0,001(mol)$
$n_{Ba(OH)_2} = 0,01.0,15 = 0,0015(mol)$

$NaOH \to Na^+ + OH^-$
$Ba(OH)_2 \to Ba^{2+} + 2OH^-$

Ta có : 

$n_{OH^-}= 0,001 + 0,0015.2 = 0,004(mol)$
$V_{dd} = 0,1 + 0,15 = 0,25(mol)$
$[OH^-] = \dfrac{0,004}{0,25} = 0,016M$

$pOH = -log(0,016) = 1,795 \Rightarrow pH = 14 - 1,795 = 12,205$

\(n_{OH^-}=0,1.0,01+0,15.0,01.2=0,004\left(mol\right)\\ V_{ddsau}=100+150=250\left(ml\right)=0,25\left(l\right)\\ \Rightarrow\left[OH^-\right]=\dfrac{0,004}{0,25}=0,016\left(M\right)\\ pH=14+log\left[0,016\right]\approx12,204\)

\(n_{NaOH}=0,03.0,1=0,003\left(mol\right)\\ n_{HNO_3}=0,01.0,01=0,0001\left(mol\right)\\ NaOH+HNO_3\rightarrow NaNO_3+H_2O\\ Vì:\dfrac{0,0001}{1}< \dfrac{0,003}{1}\\ \Rightarrow NaOHdư\\ n_{NaOH\left(dư\right)}=0,003-0,0001=0,0029\left(mol\right)\\ \left[OH^-\left(dư\right)\right]=\left[NaOH_{dư}\right]=\dfrac{0,0029}{0,01+0,1}=\dfrac{29}{1100}\left(M\right)\\ \Rightarrow pH=14+log\left[\dfrac{29}{1100}\right]\approx12,421\)

Ta có: \(\left\{{}\begin{matrix}n_{Ba^{2+}}=n_{Ba\left(OH\right)_2}=0,1.0,01=0,001\left(mol\right)\\n_{OH^-}=2n_{Ba\left(OH\right)_2}=2.0,1.0,01=0,002\left(mol\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\Sigma n_{H^+}=0,2.0,01+2.0,2.0,02=0,01\left(mol\right)\\n_{SO_4^{2-}}=n_{H_2SO_4}=0,2.0,02=0,004\left(mol\right)\end{matrix}\right.\)

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

_____ 0,01___0,002_________ (mol)

⇒ H⁺ dư. \(\Rightarrow n_{H^+\left(dư\right)}=0,008\left(mol\right)\)

\(\Rightarrow\left[H^+\right]=\frac{0,008}{0,3}=\frac{2}{75}M\Rightarrow pH\approx1,57\)

PT ion: \(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_{4\downarrow}\)

______ 0,001__0,004__ → 0,001 (mol)

\(\Rightarrow m_{\downarrow}=m_{BaSO_4}=0,001.233=0,233\left(g\right)\)

Bạn tham khảo nhé!

\(\left[H^+\right]=0,2\cdot0,01=0,002mol\\ \left[OH^-\right]=0,3\cdot0,002=0,0006mol\\ H^++OH^-\rightarrow H_2O\)

0,02 > 0,0006

\(n_{H^+dư}=0,002-0,0006=0,0014mol\\ \)

\(\Sigma_{dd}=\dfrac{0,0014}{0,5}=0,0028M\\ \Rightarrow pH\approx2,6\)

n_HCl=0,01. 0,2=2.10^-3 n_NaOH=0,002.0,3=6.10^-4

HCl --> H⁺ + Cl^- NaOH --> Na⁺ + OH ^-

2.10^-3-->2.10^-3 6.10^-4 --> 6.10^-4

H⁺ + OH ^- --> H₂O (dư H⁺)

6.10^-4 6.10^-4

=> n_H⁺ = 1,4 .10^-3

[H⁺ ] = (1,4.10^-3) / 0,5=2,8.10^-3

==> pH= -Lg(2,8.10^-3) =2,55

Em lp dưới, có thể làm sai nên bài e chưa chắc đúng đâu chị nhé!!!!