calculate the mass of KOH to be dissolved in 250cm3 of solution to give pH of 9.given molar mass of KOH=56.1?
回答 (2)
2017-12-18 11:55 pm
pH = 9
pOH = pKw - pH = 14 - 9 = 5
Then [OH⁻] = 10^-pH M = 1 × 10⁻⁵ M
No. of moles of OH⁻ in the solution = (1 × 10⁻⁵ mol/L) × (250/1000 L) = 2.5 × 10⁻⁶ mol
No. of moles of KOH to be dissolved = 2.5 × 10⁻⁶ mol
Mass of KOH to be dissolved = (2.5 × 10⁻⁶ mol) × (56.1 g/mol) = 1.4 × 10⁻⁴ g
pOH = pKw - pH = 14 - 9 = 5
Then [OH⁻] = 10^-pH M = 1 × 10⁻⁵ M
No. of moles of OH⁻ in the solution = (1 × 10⁻⁵ mol/L) × (250/1000 L) = 2.5 × 10⁻⁶ mol
No. of moles of KOH to be dissolved = 2.5 × 10⁻⁶ mol
Mass of KOH to be dissolved = (2.5 × 10⁻⁶ mol) × (56.1 g/mol) = 1.4 × 10⁻⁴ g
2017-12-18 11:51 pm
If pH of solution = 9.00
pOH = 14.00 = 9.00 = 5.00
KOH → K+ + OH- ( KOH is a strong base that dissociates completely
[OH-] = 10^-5M
The solution must be 10^-5M in KOH
250cm³ = 0.250L
Mass required = 10^-5 X 56.1 X 0.250 = 0.00014g Or 0.14mg KOH required.
pOH = 14.00 = 9.00 = 5.00
KOH → K+ + OH- ( KOH is a strong base that dissociates completely
[OH-] = 10^-5M
The solution must be 10^-5M in KOH
250cm³ = 0.250L
Mass required = 10^-5 X 56.1 X 0.250 = 0.00014g Or 0.14mg KOH required.
收錄日期: 2021-04-18 18:01:15
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