Answer:
0.086
Explanation:
A 5.2 molal aqueous solution of methyl alcohol indicates that 5.2 moles of methyl alcohol are present in 1 kilogram (or 1000 g) of water. Water has a molecular weight of 18 g/mol.
(100g)/18g/mol=55.56 mol
5.2 mol/(5.2mol+55.56 mol)=0.086
Explanation:
It is known that value of acetic acid is 4.74. And, relation between pH and is as follows.
pH = pK_{a} + log
= 4.74 + log
So, number of moles of NaOH = Volume × Molarity
= 71.0 ml × 0.760 M
= 0.05396 mol
Also, moles of = moles of
= Molarity × Volume
= 1.00 M × 1.00 L
= 1.00 mol
Hence, addition of sodium acetate in NaOH will lead to the formation of acetic acid as follows.
Initial : 1.00 mol 1.00 mol
NaoH addition: 0.05396 mol
Equilibrium : (1 - 0.05396 mol) 0 (1.00 + 0.05396 mol)
= 0.94604 mol = 1.05396 mol
As, pH = pK_{a} + log
= 4.74 + log
= 4.69
Therefore, change in pH will be calculated as follows.
pH = 4.74 - 4.69
= 0.05
Thus, we can conclude that change in pH of the given solution is 0.05.