Molarity of sodium acetate in acetate buffer Moles of acetic acid in 250.0 ml ac
ID: 1034431 • Letter: M
Question
Molarity of sodium acetate in acetate buffer
Moles of acetic acid in 250.0 ml acetate buffer
Molarity of acetic acid in acatate buffer SHOW ALLWORK! CalculationsforAcetateBuffersolutionaram? Moles of sodium acetate in 250.0 mL acetate buffer Molarity of sodium acetate in acetate buffer Moles of acetic acid in 250.0 mL acetate buffer Molarity of acetic acid in acetate buffer SHOW ALL WORK: pH Calculations (4 pts. for each blank and table) pH of 50.0 mL H,O +1.00 mL 0.6 M NaOH a.07 60+ 5Im 0ol 1 000/51.ONTM oIT) 1.93 pH of50.0 mL H2O + 1.00 mL 0.6 HCI POH :-log( 01 nr14-1015 ?17): ias 1.93 H-1H-1.3 pH of acetate buffer pH of 50.0 mL acetate buffer +1.00 mL 0.6 M NaOH Moles CHjCOOH ?? CH,COO before addition Addition after addition 67
Explanation / Answer
The molar mass of sodium acetate = 82 g/mol
Therefore, the no. of moles of sodium acetate in acetate buffer = nsodium acetate =1.3456 g/82 g mol-1 = 0.0164 mol
The molarity of sodium acetate in 250 mL acetate buffer = (0.0164 mol/250 mL) * 1000 = 0.0656 M
The no. of moles of acetic acid in acetate buffer = 3 mol/L * (3.81/1000) L = 0.01143 mol
The molarity of acetic acid in 250 mL acetate buffer = nacetic acid = (0.01143 mol/250 mL)*1000 = 0.04572 M
pH of buffer solution:
According to Henderson-Hasselbulch equaiton, pH = pKa + Log(nsodium acetate/nacetic acid)
pKa of acetic acid = 4.74
i.e. pH = 4.74 + Log(0.0164/0.01143)
i.e. pH = 4.9
pH of solution after the addition of 1 mL 0.6 M NaOH:
pH = pKa + Log(nsodium acetate + nNaOH)/(nacetic acid - nNaOH)
i.e. pH = 4.74 + Log(0.0164+0.0006)/(0.01143-0.0006)
i.e. pH = 4.94
pH of solution after the addition of 1 mL 0.6 M :
pH = pKa + Log(nsodium acetate - nHCl)/(nacetic acid + nHCl)
i.e. pH = 4.74 + Log(0.0164-0.0006)/(0.01143+0.0006)
i.e. pH = 4.86
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