{"id":197105,"date":"2025-03-06T15:26:33","date_gmt":"2025-03-06T15:26:33","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=197105"},"modified":"2025-03-06T15:26:36","modified_gmt":"2025-03-06T15:26:36","slug":"how-many-ml-of-4-50m-sodium-hydroxide-must-be-acetic-acid-solutions-to-make-a-buffer-with-phb-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/06\/how-many-ml-of-4-50m-sodium-hydroxide-must-be-acetic-acid-solutions-to-make-a-buffer-with-phb-2\/","title":{"rendered":"How many mL of 4.50M sodium hydroxide must be acetic acid solutions to make a buffer with pHb"},"content":{"rendered":"\n

How many mL of 4.50M sodium hydroxide must be acetic acid solutions to make a buffer with pHb. How many mL of 4.50 M sodium hydroxide must be added to 250.0 mL of a 0.200 Mc. Calculate the pH of a formic acid solution that contains 1.45% formic acid by mass.Question 4a. The pH of the acid and its conjugate base pairbuffer solution can be calculated by using(Assume a density of 1.01 g\/mL for the solution.)acetic acid solution to make a buffer with pH = 5.000?the Anderson Hassel Balch equation:=5.000<\/p>\n\n\n\n

The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n

To address your queries, let’s break them down into two parts:<\/p>\n\n\n\n

1. Volume of 4.50 M Sodium Hydroxide (NaOH) Needed to Prepare an Acetic Acid Buffer with pH = 5.000<\/strong><\/p>\n\n\n\n

Buffers are solutions that resist changes in pH upon the addition of small amounts of acids or bases. The pH of a buffer solution containing a weak acid (HA) and its conjugate base (A\u207b) can be calculated using the Henderson-Hasselbalch equation:<\/p>\n\n\n\n

[ \\text{pH} = \\text{p}K_a + \\log \\left( \\frac{[\\text{A}^-]}{[\\text{HA}]} \\right) ]<\/p>\n\n\n\n

For acetic acid (CH\u2083COOH), the acid dissociation constant (K\u2090) is 1.8 \u00d7 10\u207b\u2075, giving a pK\u2090:<\/p>\n\n\n\n

[ \\text{p}K_a = -\\log(1.8 \\times 10^{-5}) \\approx 4.74 ]<\/p>\n\n\n\n

Given the desired pH of 5.000:<\/p>\n\n\n\n

[ 5.000 = 4.74 + \\log \\left( \\frac{[\\text{A}^-]}{[\\text{HA}]} \\right) ]<\/p>\n\n\n\n

Solving for the ratio (\\frac{[\\text{A}^-]}{[\\text{HA}]}):<\/p>\n\n\n\n

[ \\log \\left( \\frac{[\\text{A}^-]}{[\\text{HA}]} \\right) = 5.000 – 4.74 = 0.26 ]<\/p>\n\n\n\n

[ \\frac{[\\text{A}^-]}{[\\text{HA}]} = 10^{0.26} \\approx 1.82 ]<\/p>\n\n\n\n

This ratio indicates that the concentration of acetate ions ([A\u207b]) should be 1.82 times that of acetic acid ([HA]) to achieve a pH of 5.000.<\/p>\n\n\n\n

Assuming we start with 250.0 mL of 0.200 M acetic acid:<\/p>\n\n\n\n