The pH of 0.05 M benzoic acid is 2.24. Calculate the change in pH when 1.2 g of

Question

The pH of 0.05 M benzoic acid is 2.24.  Calculate the change in pH when 1.2 g of C6H5COONa is added to 45 4-mL of 0.50 M benzoic acid, C6H5COOH. Ignore any changes in volume.  The Ka value for C6H5COOH is 6.5 x 10-5.

5-Calculate the pH of a solution that results from mixing 33 mL of 0.13 M HBrO(aq) with 36 mL of 0.13 M NaBrO(aq). The Ka value for HBrO is 2 x 10-9.

6-A buffer consists of 0.36 M NaH2PO4 and 0.41 M Na2HPO4. Given that the K values for H3PO4 are, Ka1 = 7.2 x 10-3, Ka2 = 6.3 x 10-8, and Ka3 = 4.2 x 10-13, calculate the pH for this buffer.

Explanation / Answer

1)

Given 1.2 g of C6H5C00Na of 45 ml

Moles = mass /M.W

=1.2 /122

=0.009836

Moles of C6H5C00Na = 0.009836

moles of C6H5C00H = .5 x 45 /1000= 0.0225

pKa=-log Ka

=-log 6.5 x 10-5

pKa = 4.187

pH =4.187 + log ( 0.009836/ 0.0225)

pH= 3.827

initial pH is 2.24

Change = 3.827 -2.24

Change in pH is 1.587

2) it forms a acid buffer system .

moles of HBr0 = 33 x 0.13 /1000 = 4.29 x 10-3

moles of NaBro = 36 x 0.13 /1000 = 4.68 x 10-3

pKa =-log Ka

=-log 2 x 10-9

= 8.70

pH = pKa + log ( NaBro/ HBro)

=8.70 + log ( 4.68 x 10-3 /4.29 x 10-3)

= 8.738

3) Given 0.36 M NaH2PO4and 0.41 M Na2HPO4

pH = pKa + log ( Na2HP04)/( NaH2P04)

pKa = -log ka2

=-log 6.3 x 10-8

= 7.2

pH =7.2 + log (0.41/0.36)

pH = 7.256

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