A solution household bleach contains 5.25% NaClO, by mass. Assuming that the density of bleach is the same as water(1.0g/ml). Calculate the volume of household bleach that should be diluted with water to make 500.0 ml of a ph= 10.26 solution.
So, the first thing i do was to find kb using the ka of hypochlorous acid but from there i don't know where to go. Could you help me?
You are right to calculate Kb.
ClO^- + HOH ==> HClO + OH^-
Set up an ICE chart. You will substitute as follows:
Kb = (Kw/Ka) = (HClO)(OH^-)/(ClO^-)
You know Kw and Ka. (HClO)=(OH^-) = x and (ClO^-) = 5.25% BUT that must be converted to molarity. Solve for x which is the OH^-, convert that to pOH, then to pH, then to (H^+). That is the (H^+) of the 5.25% bleach. Then use the dilution formula of
mL x M = mL x M.
Uh how do you convert the 5.25% to molarity?
wouldnt the answer be 500 ml? since the density of water is the same as the bleach?
Uh how do you convert the 5.25% to molarity?
if the solution is 5.25% then 1 litre (=1 kg) contains 52.5 g
Calculate the molecular mass for NaClO=M
then the molarity =
52.5 g/M g mole^-1
What gets substituted into the formula M1V1=M2V2? I don't understand...
Of course! Let's go through the steps together to find the volume of household bleach that should be diluted.
Step 1: Calculate the concentration of hypochlorite ions (OCl-) in the bleach solution. Since the solution contains 5.25% NaClO by mass, it means that for every 100g of bleach, there are 5.25g of NaClO. Since the density of bleach is given as 1.0g/ml and assuming the concentration is homogeneous, it means that 5.25g NaClO is present in 100 ml (or 0.1 L) of bleach.
To convert grams to moles, we need to divide the mass by the molar mass of NaClO:
Molar mass of NaClO = (1 x molar mass of Na) + molar mass of Cl + (1 x molar mass of O)
= (1 x 22.99 g/mol) + 35.45 g/mol + (1 x 16.00 g/mol)
= 22.99 g/mol + 35.45 g/mol + 16.00 g/mol
= 74.44 g/mol
Concentration of NaClO (in mol/L) = (mass of NaClO in grams) / (molar mass of NaClO in g/mol)
= 5.25 g / 74.44 g/mol
= 0.0703 mol/L
Step 2: Use the definition of pH to find the concentration of hydronium ions (H3O+):
pH = -log10[H3O+]
10^(-pH) = [H3O+]
Since the pH of the resulting solution is given as 10.26, we can calculate the concentration of H3O+:
[H3O+] = 10^(-10.26) mol/L
= 4.53 x 10^(-11) mol/L
Step 3: Write the reaction equation for the dissociation of hypochlorous acid (HClO) and find the value of the ionization constant (Ka).
HClO + H2O ⇔ H3O+ + ClO-
The value of Ka for HClO at a given temperature is obtained from a reference source. Let's assume it to be 3.5 x 10^(-8) mol/L.
Since Ka = ([H3O+][ClO-]) / [HClO], rearranging the formula, we can find [ClO-]:
[ClO-] = (Ka x [HClO]) / [H3O+]
= (3.5 x 10^(-8) x (0.0703 mol/L)) / (4.53 x 10^(-11) mol/L)
= 539.2 mol/L
Step 4: Calculate the volume of NaClO solution needed to prepare 500.0 ml of a pH = 10.26 solution.
The reaction equation between NaClO and HClO is as follows:
NaClO + HClO ⇔ Na+ + ClO- + H2O
Let V be the volume of NaClO solution needed. The concentration of ClO- in this volume is calculated using the equation:
[ClO-] = (amount of ClO- in moles) / (total volume in liters)
539.2 mol/L = (V x 0.0703 mol/L) / (V + 0.5 L)
Simplifying the equation, we find:
(V + 0.5 L) x 539.2 mol/L = V x 0.0703 mol/L
Now, solve for V:
(V + 0.5) = (0.0703 / 539.2) x V
V + 0.5 = 0.00013 x V
V = 0.5 / (1 - 0.00013)
V ≈ 0.5 / 0.99987
V ≈ 0.50013 L or 500.13 ml
Therefore, the volume of household bleach that should be diluted with water to make 500.0 ml of a pH = 10.26 solution is approximately 500.13 ml.