HELP!! 1.20g of sodium hydrogen sulphate was made into 100cm3 standard solution. 25cm3 of this solution was neutralised by 23.55cm3 NaOH. What was the concentraion of NaOH?
NaHSO4 + NaOH --> Na2SO4 + H2O
mols NaHSO4 = grams/molar mass = 1.20/ 120 = 0.01
M NaHSO4 = mols/L = 0.01/0.1 = 0.1 M
mols NaHSO4 used = M x L = 0.1M x 0.025 = 0.0025
Since 1 mol NaHSO4 = 1 mol NaOH from the balanced equation, then mols NaOH = 0.0025 and M = mols NaOH/L NaOH = 0.0025/0.02355 = 0.1062
Personally, I dislike all those zeros. It makes it easy to loose one or more of them. I prefer millimoles like this.
millimols NaHSO4 used = mL x M = 25 x 0.1 M = 2.5
millimols NaOH = 2.5
M NaOH = millimoles/mL = 2.5/23.55 = 0.1062 M
To find the concentration of NaOH, we need to first calculate the number of moles of sodium hydrogen sulphate (NaHSO4).
1. Convert the mass of sodium hydrogen sulphate to moles.
The molar mass of NaHSO4 is 22.99 g/mol for sodium (Na), 1.01 g/mol for hydrogen (H), 32.07 g/mol for sulfur (S), and 16.00 g/mol for oxygen (O).
Therefore, the molar mass of NaHSO4 is:
(1 * 22.99) + (1 * 1.01) + (1 * 32.07) + (4 * 16.00) = 120.00 g/mol
The number of moles of NaHSO4 can be calculated as follows:
moles = mass / molar mass
moles = 1.20 g / 120.00 g/mol = 0.01 mol
2. Calculate the concentration of NaHSO4 in the standard solution.
The concentration is given as the number of moles per volume, so:
concentration = moles / volume
concentration = 0.01 mol / 0.1 L (since 100 cm^3 is equal to 0.1 L) = 0.1 mol/L
3. Determine the number of moles of NaOH used in the neutralization.
Using the balanced equation for the reaction between NaHSO4 and NaOH, we can determine that the mole ratio is 1:1.
Therefore, the number of moles of NaOH used is also 0.01 mol.
4. Calculate the concentration of NaOH.
concentration = moles / volume
concentration = 0.01 mol / 0.02355 L (since 23.55 cm^3 is equal to 0.02355 L) = 0.425 mol/L
So, the concentration of NaOH is 0.425 mol/L.
To find the concentration of NaOH, we need to use the provided information about the neutralization reaction.
Let's break down the steps to solve this problem:
Step 1: Calculate the number of moles of sodium hydrogen sulfate (NaHSO₄) in the 1.20g sample.
To do this, we need to know the molar mass of NaHSO₄, which can be found by adding up the atomic masses of Na, H, S, and O:
Na: 22.99 g/mol
H: 1.01 g/mol
S: 32.07 g/mol
O: 16.00 g/mol
Molar mass of NaHSO₄ = 22.99 + 1.01 + 32.07 + (4 * 16.00) = 120.03 g/mol
Next, we can calculate the number of moles using the formula:
Number of moles = Mass / Molar mass
Number of moles of NaHSO₄ = 1.20 g / 120.03 g/mol
Step 2: Calculate the concentration of NaHSO₄ in the 100 cm³ solution.
Concentration is usually expressed in moles per liter (mol/L) or Molarity (M), so we need to convert the volume from cm³ to L:
Volume of solution = 100 cm³ / 1000 cm³/L = 0.1 L
Concentration of NaHSO₄ in the solution = Number of moles / Volume of solution
Step 3: Determine the mole ratio between NaHSO₄ and NaOH.
From the balanced chemical equation, we know that NaHSO₄ reacts with NaOH in a 1:1 ratio. This means that for every mole of NaHSO₄, we need exactly the same amount of NaOH to neutralize it.
Step 4: Calculate the number of moles of NaOH required to neutralize the NaHSO₄.
Since the mole ratio is 1:1, the number of moles of NaOH required will be the same as the number of moles of NaHSO₄.
Step 5: Calculate the concentration of NaOH using the number of moles and volume.
Concentration of NaOH = Number of moles / Volume
Given that 25 cm³ of the NaHSO₄ solution is neutralized by 23.55 cm³ of NaOH, we can conclude that the volume used for the reaction is 23.55 cm³.
Volume of NaOH = 23.55 cm³ / 1000 cm³/L = 0.02355 L
Finally, to find the concentration of NaOH:
Concentration of NaOH = Number of moles of NaOH / Volume of NaOH
Since the number of moles of NaOH required is the same as the number of moles of NaHSO₄, we can use the value calculated in Step 1.
Substitute the values into the formula to calculate the concentration of NaOH.