1)
\({n_{NaOH}} = 0,05.1 = 0,05{\text{ mol;}}{{\text{n}}_{KOH}} = 0,15.0,5 = 0,075{\text{ mol;}}{{\text{V}}_{NaOH}} = 0,05 + 0,15 = 0,2{\text{ lít}}\)
\( \to {C_{M{\text{ NaOH}}}} = \frac{{0,05}}{{0,2}} = 0,25M;{C_{M{\text{ KOH}}}} = \frac{{0,075}}{{0,2}} = 0,375M\)
2)
\({m_{KOH}} = 50.10\% = 5{\text{ gam;}}{{\text{m}}_{KCl}} = 150.15\% = 22,5{\text{ gam;}}{{\text{m}}_{dd{\text{ mới}}}} = 50 + 150 = 200{\text{ gam}}\)
\( \to C{\% _{NaOH}} = \frac{5}{{200}}.100\% = 2,5\% ;C{\% _{KCl}} = \frac{{22,5}}{{200}}.100\% = 11,25\% \)