Find the pH of a solution that is 0.231 M fromic acid and 0.183 M potassium form

Question

Find the pH of a solution that is 0.231 M fromic acid and 0.183 M potassium formate. Formic acid has a pka of 3.74 What is the pH of a solution that is 0.275 M benzoic acid and 0.129 M sodium benzoate? Benzoic acid has a pka of 4.20. what is the pH of 100 ml of a 0.205 M solution of acetic acid afer 25.0 ml. of 0.111 M NaOH have been added? Acetic acid has a pkaof 4.74 (ka = 1.8 times 10-5) what is the total alkalinity of a 100 mL sample of stream water that requires 18.53 mL of 0.0205 N HCI to bring it to the endpoint. Report the result as [HCO3-1] AND mg/L CaCO3. A 100 mL sample of well water was tested and had an initial pH of 6.67. A 100mL sample of this water required 31.08 mL of 0.0198 N HCl for complete titration. Determine the following: [HCO3-1] mg/L CaCO3 [H + ] [OH] {h2CO3] [co32] A 100mL pond sample with an initial pH of 7.16 needed 24.29 mL of 0.0210N Hcl for a complete titration. Determine the following: [HCO-13] mg/L CaCO3 [H + ] [OH] [H2CO3] [CO23]

Explanation / Answer

1. pH of acidc buffer = pKa + Log[salt/acid]

pKa = 3.74

[HCOOK] = 0.183 M [HCOOH] = 0.231 M

pH = 3.74 + log(0.183/0.231)

= 3.64

2. pH of acidc buffer = pKa + Log[salt/acid]

pKa = 4.2

[HCOOK] = 0.129 M [HCOOH] = 0.205 M

pH = 4.2 + log(0.129/0.205)

= 3.99

3. pH of acidc buffer = pKa + Log[salt/acid]

pKa = 4.2

[NaOH] = 0.111*25/125 = 0.022 M [H3COOH] = 0.205*100/125 = 0.164 M

pH = 4.74 + log(0.0222/0.164)

= 3.87

4.

No of equivalents of HCl = 18.53/1000*0.0205

     = 0.00038 equi

No of equivalents of alkalinity = 0.00038 equi

alkalinity [HCO3-] = 0.00038/0.1 = 0.0038 M

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