Question

Giúp em vs ạ

Answer 1

Câu 6:

a) 100 ml ddA có: \(\left\{{}\begin{matrix}n_{Ba\left(OH\right)_2}=0,1.0,1=0,01\left(mol\right)\\n_{NaOH}=0,1.0,1=0,01\left(mol\right)\end{matrix}\right.\)

=> \(n_{OH^-}=0,01+0,01.2=0,03\left(mol\right)\)

=> \(\left[OH^-\right]=\dfrac{0,03}{0,1}=0,3M\)

=> \(pH=14+\text{log}\left(0,3\right)=13,477\)

400 ml dd B có: \(\left\{{}\begin{matrix}n_{H_2SO_4}=0,4.0,0375=0,015\left(mol\right)\\n_{HCl}=0,4.0,0125=0,005\left(mol\right)\end{matrix}\right.\)

=> \(n_{H^+}=0,015.2+0,005=0,035\left(mol\right)\)

=> \(\left[H^+\right]=\dfrac{0,035}{0,4}=0,0875M\)

=> \(pH=-\text{log}\left(0,0875\right)=1,058\)

b) Ta có: \(\left\{{}\begin{matrix}n_{Ba^{2+}}=n_{Ba\left(OH\right)_2}=0,01\left(mol\right)\\n_{SO_4^{2-}}=n_{H_2SO_4}=0,015\left(mol\right)\end{matrix}\right.\)

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

                   0,03-->0,03

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

                   0,01---->0,01------>0,01

=> \(n_{H^+\left(dư\right)}=\dfrac{0,015-0,01}{0,4+0,1}=0,01\)

=> \(pH=-\text{log}\left(0,01\right)=2\)

m_{kết tủa} = m_BaSO4 = 0,01.233 = 2,33 (g)