Đáp án:
Bạn tham khảo lời giải nhé!!
Giải thích các bước giải:
b,
\(\begin{array}{l}
HN{O_3} + NaOH \to NaN{O_3} + {H_2}O\\
{n_{HN{O_3}}} = 0,001mol\\
{n_{NaOH}} = 0,001mol\\
\to {n_{HN{O_3}}} = {n_{NaOH}} = 0,001mol
\end{array}\)
Suy ra phản ứng vừa đủ
\(\begin{array}{l}
\to {n_{N{a^ + }}} = {n_{N{O_3}^ - }} = 0,001mol\\
\to C{M_{N{a^ + }}} = C{M_{N{O_3}^ - }} = \dfrac{{0,001}}{{0,2}} = 0,005M
\end{array}\)
c,
\(\begin{array}{l}
{n_{Ba{{(OH)}_2}}} = 0,001mol\\
\to {n_{B{a^{2 + }}}} = {n_{Ba{{(OH)}_2}}} = 0,001mol\\
\to {n_{O{H^ - }}} = 2{n_{Ba{{(OH)}_2}}} = 0,002mol\\
\to C{M_{B{a^{2 + }}}} = \dfrac{{0,001}}{{0,1}} = 0,01M\\
\to C{M_{O{H^ - }}} = \dfrac{{0,002}}{{0,1}} = 0,02M
\end{array}\)
d,
\(\begin{array}{l}
{n_{NaCl}} = 0,0003mol\\
{n_{NaOH}} = 0,00005mol\\
\to {n_{N{a^ + }}} = {n_{NaCl}} + {n_{NaOH}} = 0,00035mol\\
\to {n_{C{l^ - }}} = {n_{NaCl}} = 0,0003mol\\
\to {n_{O{H^ - }}} = {n_{NaOH}} = 0,00005mol\\
\to C{M_{N{a^ + }}} = \dfrac{{0,00035}}{{0,35}} = 0,001M\\
\to C{M_{C{l^ - }}} = \dfrac{{0,0003}}{{0,35}} = 0,0008M\\
\to C{M_{O{H^ - }}} = \dfrac{{0,00005}}{{0,35}} = 0,0001M
\end{array}\)
e,
\(\begin{array}{l}
{n_{HCl}} = 0,001mol\\
{n_{NaCl}} = 0,0005mol\\
\to {n_{N{a^ + }}} = {n_{NaCl}} = 0,0005mol\\
\to {n_{{H^ + }}} = {n_{HCl}} = 0,001mol\\
\to {n_{C{l^ - }}} = {n_{HCl}} + {n_{NaCl}} = 0,0015mol\\
\to C{M_{N{a^ + }}} = \dfrac{{0,0005}}{{0,15}} = 0,0033M\\
\to C{M_{C{l^ - }}} = \dfrac{{0,0015}}{{0,15}} = 0,01M\\
\to C{M_{{H^ + }}} = \dfrac{{0,001}}{{0,15}} = 0,0067M
\end{array}\)
g, Bạn bổ sung đề dùm mình nhé!!!!
