Question

Cho m (g) ba vào nước thu được 100ml dd ba(oh)2 có ph =12 tìm m

Answer 1

$[OH^-] = 10^{-14} : 10^{-pH} = 10^{-2}M$

$n_{OH^-} = 10^{-2}.0,1 = 0,001(mol)$

$n_{Ba} = n_{Ba(OH)_2} = \dfrac{1}{2}n_{OH^-} = 0,0005(mol)$

$m = 0,0005.137 = 0,0685(gam)$

Answer 2

Có : \(n_{Ba}=\dfrac{m}{M}=\dfrac{m}{137}\left(mol\right)\)

\(BTNT\Rightarrow n_{Ba\left(OH\right)2}=\dfrac{m}{137}\left(mol\right)\)

\(\Rightarrow n_{OH^-}=\dfrac{2m}{137}\left(mol\right)\)

\(\Rightarrow\left[OH^-\right]=\dfrac{20m}{137}M\)

Mà \(pH=12\Rightarrow pOH=2\Rightarrow\left[OH^-\right]=10^{-2}\)

\(\Rightarrow m=0,0685\left(g\right)\)

Vậy ..