a,

$n_{NaOH(1)}=0,4.0,5=0,2(mol)$

$m_{dd}=200.1,25=250g$

$\to n_{NaOH(2)}=\dfrac{250.20\%}{40}=1,25(mol)$

Sau khi trộn:

$V_{dd}=0,4+0,2=0,6l$

$n_{NaOH}=1,25+0,2=1,45(mol)$

$\to C_{M_{NaOH}}=\dfrac{1,45}{0,6}=\dfrac{29}{12}M$

$\to [Na^+]=[OH^-]=\dfrac{29}{12}M$

b,

$n_{CaCl_2(1)}=0,15.0,5=0,075(mol)$

$n_{CaCl_2(2)}=0,05.2=0,1(mol)$

Sau khi trộn:

$n_{CaCl_2}=0,1+0,075=0,175(mol)$

$V_{dd}=0,15+0,05=0,2l$

$\to C_{M_{CaCl_2}}=\dfrac{0,175}{0,2}=0,875M$

$\to [Ca^{2+}]=0,875M; [Cl^-]=0,875.2=1,75M$

 

1.

nNaOH=0,4.0,5=0,2mol

mdd NaOH=1,25.200=250g

mNaOH=250.20/100=50g

nNaOH=50/40=1,25mol

Tổng nNaOH=0,2+1,25=1,45mol

Vdd sau pứ=0,4+0,2=0,6l

CM NaOH=1,45/0,6=2,42M

2.

nCaCl2=0,15.0,5=0,075mol

nCaCl2=0,05.2=0,1mol

Tổng nCaCl2=0,1+0,075=0,175mol

Vdd sau pứ=0,15+0,05=0,2l
CM CaCl2=0,175/0,2=0,875M