At 90,^oC,pure water has [H^{+}] = 10^{-6},M. | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Step $1$: Calculate the millimoles of $HCl$ and $KOH$.Millimoles of $HCl = 100\, mL 	imes 0.2\, M = 20\, mmol$.Millimoles of $KOH = 200\, mL 	im

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) Step $1$: Calculate the millimoles of $HCl$ and $KOH$.Millimoles of $HCl = 100\, mL 	imes 0.2\, M = 20\, mmol$.Millimoles of $KOH = 200\, mL 	imes 0.1\, M = 20\, mmol$.Step $2$: Determine the nature of the resulting solution.Since the millimoles of $HCl$ (acid) equal the millimoles of $KOH$ (base),the reaction $HCl + KOH ightarrow KCl + H_2O$ results in a neutral solution.Step $3$: Calculate the $pH$ of the neutral solution at $90\,^oC$.At $90\,^oC$,the concentration of $[H^{+}]$ in pure water is $10^{-6}\,M$.For a neutral solution,$[H^{+}] = [OH^{-}] = 10^{-6}\,M$.$pH = -\log[H^{+}] = -\log(10^{-6}) = 6$.

At $90\,^oC$,pure water has $[H^{+}] = 10^{-6}\,M$. If $100\, mL$ of $0.2\, M\, HCl$ is added to $200\, mL$ of $0.1\, M\, KOH$ at $90\,^oC$,then the $pH$ of the resulting solution will be:

Similar Questions

When $NaCl$ is dissolved in water,the sodium ion becomes

At $80\,^{\circ}C,$ distilled water has $[H_3O^{+}]$ concentration equal to $1 \times 10^{-6} \, mol/L.$ The value of $K_w$ at this temperature will be

$100 \ mL$ of $NaOH$ solution has a $pH = 10$. Calculate the concentration of hydroxide ions,$[OH^{-}]$.

Which of the following will have the highest $pH$?

The concentration of hydrogen ion in a sample of soft drink is $3.8 \times 10^{-3} \ M$. What is its $pH$ (in $.42$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module