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calculate the pH of a solution that results from the addition of the following amounts of...

calculate the pH of a solution that results from the addition of the following amounts of 0.1 M NaOH to 10mL of a 0.1 HCl solution. Find the pHs for the following volumes of 0.1M NaOH added: 1, 2, 4,8, 9.8, 10.1, 10.2, 10.4, 11, 14, 18, 20 (all in mL)

Expert Solution

NaOH + HCl --------------> NaCl +H2O

0.1xV 10x0.1 0 0

When V =1mL

0 0.9 0.1 - thus [H+] = 0.9/11 pH = -log0.9/11 =1.087

when V = 2 mL

0 0.8 0.2 - thus [H+] = 0.8/12 pH = -log0.8/12=1.1760

Wne V = 4 mL

0 0.6 0.4 - thus [H+] = 0.6/14 pH = -log0.6/14 =1.3679

When V = 8mL

0 0.2 0.8 - thus [H+] = 0.2/18 pH = -log0.2/18 =1.9542

When V = 9.8

0 0.02 0.98 - thus [H+] = 0.02/19.8 pH = -log0.02/19.8 =2.9956

When V = 10.1 mL

0.01 0 1.00 - thus [OH-] = 0.01/20.1 pH = 14 -[-log0.01/20.1] =10.6968

When V = 10.2 mL

0.02 0 1.00 -thus [OH-] = 0.02/20.2 pH = 14 -[-log0.02/20.2] =10.9956

When V = 10.4mL

0.04 0 1.00 - thus [OH-] = 0.04/20.4 pH = 14 -[-log0.04/20.4] = 11.2924

When V = 11

0.1 0 1.00 -thus [OH-] = 0.1/21 pH = 14 -[-log0.1/21] =11.6777

When V = 14mL

0.4 0 1.00 - thus [OH-] = 0.4/24 pH = 14 -[-log0.4/24] =12.2218

When V= 18 mL

0.8 0 1.00 -thus [OH-] = 0.8/28 pH = 14 -[-log0.8/28] =12.4559

Whe V = 20mL

1.0 0 1.00 -thus [OH-] = 1.0/30 pH = 14 -[-log1.0/30] = 12.5228

queen_honey_blossom answered 2 years ago

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