the ph of the resulting solution when 1.15g of sodium metal is reacted slowly in 500cm3 of water is close to
Rxn: Na⁰(s) + H₂O(l) => NaOH(aq) + ½H₂(g)
Moles Na⁰(s) = (1.50 gms / 23 gms/mol) = 0.065 mol Na⁰(s)
From Rxn ratios …
Moles Na⁰(s) consumed = Moles NaOH formed
=> 0.065 mol Na⁰(s) => 0.065 mol NaOH(aq)
=> [NaOH] = (0.065 mole NaOH / 0.500 Liter Solution) = 0.13M NaOH
=> Since NaOH is a strong Gp-IA Base, it will ionize 100% => 0.13M Na⁺(aq) + 0.13M OH¯(aq)
=> pOH = -log[OH¯(aq)] = -log(0.13) = 0.886
= > pH + pOH = 14 => pH = 14 – pOH = 14 – 0.886 = 13.11
To determine the pH of the resulting solution when 1.15g of sodium metal is reacted slowly in 500cm3 of water, we need to understand the reaction that occurs.
When sodium metal reacts with water, it produces sodium hydroxide (NaOH) and hydrogen gas (H2). The balanced chemical equation for this reaction is:
2Na + 2H2O -> 2NaOH + H2
In this reaction, 2 moles of sodium react with 2 moles of water to produce 2 moles of sodium hydroxide and 1 mole of hydrogen gas.
To find the moles of sodium used, we can use the molar mass of sodium, which is 22.99 g/mol. So, the number of moles of sodium is calculated as:
moles of sodium = mass of sodium / molar mass of sodium
= 1.15g / 22.99 g/mol
To find the resulting concentration of sodium hydroxide in the solution, we need to convert the volume of water to liters. Since 1 cm3 is equal to 1 mL, the volume of water is 500 mL, which is equivalent to 0.5 L.
The concentration (in moles per liter) can then be calculated as:
concentration of NaOH = moles of NaOH / volume of water (in liters)
= 2 moles / 0.5 L
Finally, to determine the pH of the resulting solution, we need to calculate the hydroxide ion concentration (OH-) from the concentration of sodium hydroxide. Since sodium hydroxide is a strong base, it dissociates completely in water, meaning that for every mole of NaOH, there will be one mole of hydroxide ions (OH-).
Therefore, the hydroxide ion concentration is equal to the concentration of sodium hydroxide.
Now, to find the pH of the solution, we can use the equation:
pOH = -log10[OH-]
Since pH + pOH = 14 (at 25°C), we can rearrange the equation to find the pH:
pH = 14 - pOH
By plugging in the hydroxide ion concentration (OH-) into this equation, you can calculate the pH of the resulting solution when 1.15g of sodium metal is reacted slowly in 500cm3 of water.
Sorry, used 1.50 gms Na instead of 1.15 gms Na. Just change the 1.5 to 1.15 and the problem is the same.
[Na⁰(s)]added = [NaOH]formed = [OH¯]ionized = 1.15g/23g-mol¯¹ = 0.05 M OH¯
pOH = -log[OH¯] = -log(0.05) = 1.3 => pH = 14 – pOH = 14 – 1.3 = 12.7