At 25 °C, how many dissociated h ions are there in 341 ml of an aqueous solution whose ph is 11. 93?
1 answer:
0.017× dissociated H ions are there in 341 ml of an aqueous solution whose ph is 11. 93
If pH = 11.93
pOH = 14.00 - 11.93 = 2.07
[OH- ] =
[OH-] = 0.00851138 M
In 1 L solution we have 0.00851138 M moles OH- ions
In 0.341 L we have,
0.00851138 * 0.341
=0.0029 moles OH- ions
1 mol OH- ions = 6.022*10²³ ions
0.0029 moles OH- ions = ( 0.0029) * (6.022*10²³)
= 0.017×
Therefore, 0.017× dissociated H ions are there in 341 ml of an aqueous solution whose ph is 11. 93
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Answer:
0.1357 M
Explanation:
(a) The balanced reaction is shown below as:
(b) Moles of can be calculated as:
Or,
Given :
For :
Molarity = 0.1450 M
Volume = 10.00 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 10×10⁻³ L
Thus, moles of :
Moles of = 0.00145 moles
From the reaction,
1 mole of react with 2 moles of NaOH
0.00145 mole of react with 2*0.00145 mole of NaOH
Moles of NaOH = 0.0029 moles
Volume = 21.37 mL = 21.37×10⁻³ L
Molarity = Moles / Volume = 0.0029 / 21.37×10⁻³ M = 0.1357 M