What tool would you use to test for hardness?

A chemistry student weighs out 0.104 g of sulfurous acid, a diprotic acid, into a 250.0 mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.0700 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the final equivalence point. Be sure your answer has the correct number of significant digits.

Answer: The volume of NaOH needed is 36.2 mL

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of sulfurous acid = 0.104 g

Molar mass of sulfurous acid = 82 g/mol

Volume of solution = 250 mL

Putting values in above equation, we get:

$\text{Molarity of sulfurous acid}=\frac{0.104\times 1000}{82\times 250}\\\\\text{Molarity of sulfurous acid}=5.07\times 10^{-3}M$

To calculate the volume of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_3$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=5.07\times 10^{-3}M\\V_1=250mL\\n_2=1\\M_2=0.0700M\\V_2=?mL$

Putting values in above equation, we get:

$2\times 5.07\times 10^{-3}\times 250.0=1\times 0.0700\times V_2\\\\V_2=\frac{2\times 5.07\times 10^{-3}\times 250}{1\times 0.0700}=36.2mL$

Hence, the volume of NaOH needed is 36.2 mL

5 0

3 years ago

How to find an experimental error

gulaghasi [49]

Divide the difference by the accepted value and Multiply times 100 to make the value a percent. Use significant figures in all your calculations. When you subtract (Step #1) round your answer to the correct number of significant figures.

6 0

3 years ago

Given 5.0 grams of lead (II) nitrate and 3.0

natulia [17]

Answer:

1.70 g of NaNO₃

Explanation:

The balance chemical equation for given double displacement reaction is,

Pb(NO₃)₂ + 2 NaI → PbI₂ + 2 NaNO₃

Step 1: Calculate moles of each reactant:

Moles = Mass / M.Mass

For Pb(NO₃)₂:

Moles = 5.0 g / 331.21 g/mol

Moles = 0.0150 moles

For NaI:

Moles = 3.0 g / 149.89 g/mol

Moles = 0.020 moles

Step 2: Calculate Limiting reagent as;

According to equation,

1 mole of Pb(NO₃)₂ reacts with = 2 moles of NaI

So,

0.0150 moles of Pb(NO₃)₂ will react with = X moles of NaI

Solving for X,

X = 2 mol × 0.0150 mol / 1 mol

X = 0.030 moles of NaI

Hence, it means that NaI is the limiting reagent therefore, it will control the yield of Sodium Nitrate.

Step 3: Find out moles of NaNO₃ formed:

According to equation,

2 mole of NaI produced = 2 moles of NaNO₃

So,

0.020 moles of NaI will produce = X moles of NaNO₃

Solving for X,

X = 2 mol × 0.020 mol / 2 mol

X = 0.020 moles of NaNO₃

Step 4: Calculate Mass of NaNO₃ produced;

As, Moles = Mass / M.Mass

Or,

Mass = Moles × M.Mass

Putting values,

Mass = 0.020 mol × 84.99 g/mol

Mass = 1.70 g of NaNO₃

7 0

3 years ago