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a) \(n_{H^+}=n_{HCl}+2n_{H_2SO_4}=0,1\cdot0,12+0,1\cdot0,04=0,016\)
\(C_M=\dfrac{0,016}{0,2}=0,08M\)
\(\Rightarrow pH=-log\left(0,08\right)=1,1\)
b) \(n_{OH^-}=n_{KOH}+2n_{Ba\left(OH\right)_2}=0,012+2\cdot0,004=0,02\)
\(C_M=\dfrac{0,02}{0,2}=0,1\)
\(\Rightarrow pH=-log\left(\dfrac{10^{-14}}{0,1}\right)=13\)
CNaOH sau = 0,01*100/(100+100)=0.005M
CKOH sau= 0,02*100/(100+100)=0,01M
NaOH →Na+ + OH-
0,005---------->0,005 (M)
KOH →K+ + OH-
0,01--------->0,01(M)
=> [OH-]=0.01+0.005=0.015=> [H+]=10-14:0,015=6,67.10-13 (M)
=> pH= -log(6,67.10-13)= 12,18
a, \(n_{H^+}=0,025.0,2=0,005\left(mol\right)\)
\(n_{OH^-}=0,01.2.0,3=0,006\left(mol\right)\)
\(\Rightarrow n_{OH^-dư}=0,001\left(mol\right)\)
\(\Rightarrow\left[OH^-\right]_{dư}=\dfrac{0,001}{1}=10^{-3}\)
\(\Rightarrow\left[H^+\right]=10^{-11}\)
\(\Rightarrow pH=11\)
b, \(n_{Fe^{2+}}=n_{SO_4^{2-}}=0,02.0,1=0,002\left(mol\right)\)
\(n_{Ba^{2+}}=0,01.0,3=0,003\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{BaSO_4\downarrow}=n_{SO_4^{2-}}=0,002\left(mol\right)\\n_{Fe\left(OH\right)_2\downarrow}=n_{OH^-dư}=0,001\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{\downarrow}=0,002.233+0,001.90=0,556\left(g\right)\)
Ta có: \(n_{HCl}=0,1.0,02=0,002\left(mol\right)\Rightarrow n_{H^+}=0,002\left(mol\right)\)
\(n_{NaOH}=0,1.0,04=0,004\left(mol\right)\Rightarrow n_{OH^-}=0,004\left(mol\right)\)
\(H^++OH^-\rightarrow H_2O\)
0,002→0,002 _______ (mol)
⇒ OH- dư.
\(\Rightarrow n_{OH^-\left(dư\right)}=0,002\left(mol\right)\)
\(\Rightarrow\left[OH^-\right]=\frac{0,002}{0,2}=0,01M\)
\(\Rightarrow\left[H^+\right]=\frac{10^{-14}}{0,01}=10^{-12}\)
\(\Rightarrow pH=12\)
Bạn tham khảo nhé!