7.1 A standard solution is prepared by dissolving 10 g of sodium carbonate, Na2CO3(s), in 0.7 dm³ of water - English General - NSC Physical Sciences - Question 7 - 2024 - Paper 2
Question 7
7.1 A standard solution is prepared by dissolving 10 g of sodium carbonate, Na2CO3(s), in 0.7 dm³ of water.
7.1.1 Calculate the concentration of the solution.
7.1.... show full transcript
Worked Solution & Example Answer:7.1 A standard solution is prepared by dissolving 10 g of sodium carbonate, Na2CO3(s), in 0.7 dm³ of water - English General - NSC Physical Sciences - Question 7 - 2024 - Paper 2
Step 1
7.1.1 Calculate the concentration of the solution.
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Answer
To calculate the concentration of the sodium carbonate solution, we use the formula:
C=M×Vm
Where:
m = mass of sodium carbonate = 10 g
M = molar mass of Na2CO3 = 106 g/mol
V = volume of the solution = 0.7 dm³
Substituting the values, we get:
C=106×0.710=0.13 mol/dm3
Step 2
7.1.2 Will the pH of the solution be GREATER THAN or LESS THAN 7?
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The pH of the sodium carbonate solution will be GREATER THAN 7.
This is because sodium carbonate is a salt derived from a weak acid (carbonic acid) and a strong base (sodium hydroxide), leading to basic conditions in the solution.
Step 3
7.1.3 Write an equation that explains the answer to QUESTION 7.1.2.
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The dissociation of sodium carbonate in water can be represented as follows:
Na2CO3(aq)+H2O(l)⇌2Na+(aq)+HCO3−(aq)+OH−(aq)
Here, the generation of hydroxide ions (OH−) confirms that the solution is basic, leading to a pH greater than 7.
Step 4
7.1.4 Which ONE of the indicators (P, Q or R) is most suitable for this titration?
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The most suitable indicator for the titration of sodium carbonate with hydrochloric acid is indicator P, which has a pH range of 3.4 to 4.5.
This choice is appropriate because the equivalence point of the titration will be below pH 7 due to the weak base (sodium carbonate) reacting with a strong acid (hydrochloric acid).
Step 5
7.2.1 What is meant by a dilute acid?
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A dilute acid is an acid that has a relatively low concentration of hydrogen ions (H+) compared to a concentrated acid. This means that the acid is mixed with a significant amount of water, resulting in a solution where the pH is generally higher than that of a concentrated acid.
Step 6
7.2.2 Calculate the pH of the final solution.
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First, we calculate the moles of [H+] ions:
Since H2SO4 is a diprotic acid, it produces 2 moles of H+ for every mole:
Moles of H+ from H2SO4: 0.01extmol×2=0.02extmol.
Moles of OH− from KOH: 0.024 mol.
Since OH− exceeds H+, we calculate the excess moles:
Excess OH−: 0.024−0.02=0.004extmol in 0.2 dm³.
Concentration of OH−:
[OH−]=0.20.004=0.02extmol/dm3
Using the relation:
pOH=−log[OH−]=−log(0.02)≈1.7
Then, we find the pH using:
pH+pOH=14⇒pH=14−1.7=12.3