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lesya692 [45]
10 months ago
7

a 1.513 g sample of khp (c8h5o4k) is dissolved in 50.0 ml of di water. when the khp solution was titrated with naoh, 14.8 ml was

required to reach the phenolphthalien end point. calculate the molarity of the naoh solution? khp is a monoprotic acid.
Chemistry
1 answer:
JulijaS [17]10 months ago
4 0

molar mass of KHP ( C8H5O4K)= 8*12+5+4*16+39=204

( atomic weights : C=12, H=1, K=39 and O= 16)

moles of KHP= mass/molar mass =  1.513/204 =0.0074

The reaction between KHP and NaOH is

KHC8H4O4(aq) + NaOH(aq) => KNaC8H4O4(aq) + H2O(l)

1 mole of KHP requires 1 mole of NaOH;

moles of KHP=0.0074

moles of NaOH= 0.0074

concentration of NaOH= moles/ Volume in L, 1000ml= 1L, 14.8ml= 14.8/1000L=0.0148L

concentration of NaOH=0.0074/0.0148=0.5M

The molarity of NaOH is 0.5M

Molar Mass:

In chemistry, the molar mass of a compound is defined as the mass of a sample of that compound divided by the amount of substance. This is the number of moles of that sample measured in moles.

Molarity:

Molarity is used to describe the concentration of a solution. Molarity, also called molarity, is the number of moles of solute (dissolved substance) per liter of solution.

Learn more about molarity:

brainly.com/question/19943363

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Oxalic acid can remove rust (Fe2O3) caused by bathtub rings according to the reaction Fe2O3(s) - 6H2C2O4(aq) rightarrow 2Fe(C2O4
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<u>Answer:</u> The mass of rust that can be removed is 1.597 grams

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .....(1)

Molarity of oxalic acid solution = 0.1255 M

Volume of solution = 6.00\times 10^2mL = 600 mL = 0.600 L    (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

0.100M=\frac{\text{Moles of oxalic acid}}{0.600L}\\\\\text{Moles of oxalic acid}=(0.100mol/L\times 0.600L)=0.06mol

For the given chemical reaction:

Fe_2O_3(s)+6H_2C_2O_4(aq.)\rightarrow 2Fe(C_2O_4)_3^{3-}(aq.)+3H_2O(l)+6H^+(aq.)

By Stoichiometry of the reaction:

6 moles of oxalic acid reacts with 1 mole of ferric oxide (rust)

So, 0.06 moles of oxalic acid will react with = \frac{1}{6}\times 0.06=0.01mol of ferric oxide (rust)

To calculate the mass of rust for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of rust (ferric oxide) = 159.7 g/mol

Moles of rust = 0.01 moles

Putting values in above equation, we get:

0.01mol=\frac{\text{Mass of rust}}{159.7g/mol}\\\\\text{Mass of rust}=(0.01mol\times 159.7g/mol)=1.597g

Hence, the mass of rust that can be removed is 1.597 grams

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