315023
\(\mathrm{pH}\) of a sample of \(\mathrm{KOH}\) and another sample of \(\mathrm{NaOH}\) are 10 and 12 respectively. Their normalities are related as \(\mathrm{N}_{\mathrm{NaOH}}=\mathrm{xN}_{\mathrm{KOH}}\). What is the value of \({\text{x}}\) ?
1 \(5 / 6\)
2 \(6 / 5\)
3 \(10^{2}\)
4 \(10^{-2}\)
Explanation:
\(\mathrm{pH}\) of \(\mathrm{KOH}=10 ; \mathrm{pH}\) of \(\mathrm{NaOH}=12\) So, \(\mathrm{pOH}\) of \(\mathrm{KOH}=14-10=4\); \(\mathrm{pOH}\) of \(\mathrm{NaOH}=14-12=2\). \(\mathrm{N}_{\text {кон }}=\left[\mathrm{OH}^{-}\right]=10^{-4}\); \(\mathrm{N}_{\mathrm{NaOH}}=10^{-2}\) (Because, for strong base, \(\left[\mathrm{OH}^{-}\right]=\)normality \(\left.=\mathrm{pOH}\right)\) As per the question, \(10^{-2}=\mathrm{x} \times 10^{-4}\) \(\Rightarrow \mathrm{x}=\dfrac{10^{-2}}{10^{-4}}=10^{2}\)
CHXI07:EQUILIBRIUM
315025
\({\text{10 mL}}\) of \({\text{0}}{\text{.1 N HCl}}\) is added to \({\text{ 990 mL}}\) solution of \(\mathrm{NaCl}\), the \(\mathrm{pH}\) of resulting solution is
315023
\(\mathrm{pH}\) of a sample of \(\mathrm{KOH}\) and another sample of \(\mathrm{NaOH}\) are 10 and 12 respectively. Their normalities are related as \(\mathrm{N}_{\mathrm{NaOH}}=\mathrm{xN}_{\mathrm{KOH}}\). What is the value of \({\text{x}}\) ?
1 \(5 / 6\)
2 \(6 / 5\)
3 \(10^{2}\)
4 \(10^{-2}\)
Explanation:
\(\mathrm{pH}\) of \(\mathrm{KOH}=10 ; \mathrm{pH}\) of \(\mathrm{NaOH}=12\) So, \(\mathrm{pOH}\) of \(\mathrm{KOH}=14-10=4\); \(\mathrm{pOH}\) of \(\mathrm{NaOH}=14-12=2\). \(\mathrm{N}_{\text {кон }}=\left[\mathrm{OH}^{-}\right]=10^{-4}\); \(\mathrm{N}_{\mathrm{NaOH}}=10^{-2}\) (Because, for strong base, \(\left[\mathrm{OH}^{-}\right]=\)normality \(\left.=\mathrm{pOH}\right)\) As per the question, \(10^{-2}=\mathrm{x} \times 10^{-4}\) \(\Rightarrow \mathrm{x}=\dfrac{10^{-2}}{10^{-4}}=10^{2}\)
CHXI07:EQUILIBRIUM
315025
\({\text{10 mL}}\) of \({\text{0}}{\text{.1 N HCl}}\) is added to \({\text{ 990 mL}}\) solution of \(\mathrm{NaCl}\), the \(\mathrm{pH}\) of resulting solution is
315023
\(\mathrm{pH}\) of a sample of \(\mathrm{KOH}\) and another sample of \(\mathrm{NaOH}\) are 10 and 12 respectively. Their normalities are related as \(\mathrm{N}_{\mathrm{NaOH}}=\mathrm{xN}_{\mathrm{KOH}}\). What is the value of \({\text{x}}\) ?
1 \(5 / 6\)
2 \(6 / 5\)
3 \(10^{2}\)
4 \(10^{-2}\)
Explanation:
\(\mathrm{pH}\) of \(\mathrm{KOH}=10 ; \mathrm{pH}\) of \(\mathrm{NaOH}=12\) So, \(\mathrm{pOH}\) of \(\mathrm{KOH}=14-10=4\); \(\mathrm{pOH}\) of \(\mathrm{NaOH}=14-12=2\). \(\mathrm{N}_{\text {кон }}=\left[\mathrm{OH}^{-}\right]=10^{-4}\); \(\mathrm{N}_{\mathrm{NaOH}}=10^{-2}\) (Because, for strong base, \(\left[\mathrm{OH}^{-}\right]=\)normality \(\left.=\mathrm{pOH}\right)\) As per the question, \(10^{-2}=\mathrm{x} \times 10^{-4}\) \(\Rightarrow \mathrm{x}=\dfrac{10^{-2}}{10^{-4}}=10^{2}\)
CHXI07:EQUILIBRIUM
315025
\({\text{10 mL}}\) of \({\text{0}}{\text{.1 N HCl}}\) is added to \({\text{ 990 mL}}\) solution of \(\mathrm{NaCl}\), the \(\mathrm{pH}\) of resulting solution is
315023
\(\mathrm{pH}\) of a sample of \(\mathrm{KOH}\) and another sample of \(\mathrm{NaOH}\) are 10 and 12 respectively. Their normalities are related as \(\mathrm{N}_{\mathrm{NaOH}}=\mathrm{xN}_{\mathrm{KOH}}\). What is the value of \({\text{x}}\) ?
1 \(5 / 6\)
2 \(6 / 5\)
3 \(10^{2}\)
4 \(10^{-2}\)
Explanation:
\(\mathrm{pH}\) of \(\mathrm{KOH}=10 ; \mathrm{pH}\) of \(\mathrm{NaOH}=12\) So, \(\mathrm{pOH}\) of \(\mathrm{KOH}=14-10=4\); \(\mathrm{pOH}\) of \(\mathrm{NaOH}=14-12=2\). \(\mathrm{N}_{\text {кон }}=\left[\mathrm{OH}^{-}\right]=10^{-4}\); \(\mathrm{N}_{\mathrm{NaOH}}=10^{-2}\) (Because, for strong base, \(\left[\mathrm{OH}^{-}\right]=\)normality \(\left.=\mathrm{pOH}\right)\) As per the question, \(10^{-2}=\mathrm{x} \times 10^{-4}\) \(\Rightarrow \mathrm{x}=\dfrac{10^{-2}}{10^{-4}}=10^{2}\)
CHXI07:EQUILIBRIUM
315025
\({\text{10 mL}}\) of \({\text{0}}{\text{.1 N HCl}}\) is added to \({\text{ 990 mL}}\) solution of \(\mathrm{NaCl}\), the \(\mathrm{pH}\) of resulting solution is