Đáp án:

 a) 0,1875l

b) 0,86M

c) 150g

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
KOH + HCl \to KCl + {H_2}O\\
{n_{HCl}} = {C_M} \times V = 0,25 \times 1,5 = 0,375\,mol\\
{n_{KOH}} = {n_{HCl}} = 0,375\,mol\\
{V_{KOH}} = \dfrac{n}{{{C_M}}} = \dfrac{{0,375}}{2} = 0,1875l\\
b)\\
{V_{{\rm{dd}}spu}} = 0,1875 + 0,25 = 0,4375l\\
{n_{KCl}} = {n_{HCl}} = 0,375\,ml\\
{C_M}KCl = \dfrac{n}{V} = \dfrac{{0,375}}{{0,4375}} = 0,86M\\
c)\\
NaOH + HCl \to NaCl + {H_2}O\\
{n_{NaOH}} = {n_{HCl}} = 0,375\,mol\\
{m_{NaOH}} = n \times M = 0,375 \times 40 = 15g\\
{m_{{\rm{dd}}NaOH}} = \dfrac{{{m_{NaOH}} \times 100}}{{{C_\% }}} = \dfrac{{15 \times 100}}{{10}} = 150g
\end{array}\)