{"id":233026,"date":"2025-06-12T13:19:20","date_gmt":"2025-06-12T13:19:20","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=233026"},"modified":"2025-06-12T13:19:22","modified_gmt":"2025-06-12T13:19:22","slug":"sodium-benzoateis-a-product-of-sodium-hydroxide-and-benzoic-acid","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/12\/sodium-benzoateis-a-product-of-sodium-hydroxide-and-benzoic-acid\/","title":{"rendered":"Sodium benzoateis a product of sodium hydroxide and benzoic acid"},"content":{"rendered":"\n

Sodium benzoate
is a product of sodium hydroxide and benzoic acid
(
). What is the pH of 0.12 M Sodium benzoate? Please enter 3 significant figures (Sample problem 18.12 and follow-up problems)<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Answer:<\/strong> 8.64<\/p>\n\n\n\n

Explanation:<\/h3>\n\n\n\n

Sodium benzoate (C\u2086H\u2085COONa) is the salt formed from a strong base (sodium hydroxide, NaOH) and a weak acid (benzoic acid, C\u2086H\u2085COOH). When dissolved in water, it dissociates completely into its constituent ions:<\/p>\n\n\n\n

C\u2086H\u2085COONa(aq) \u2192 Na\u207a(aq) + C\u2086H\u2085COO\u207b(aq)<\/p>\n\n\n\n

The sodium ion (Na\u207a) is the conjugate acid of a strong base and acts as a spectator ion, meaning it does not react with water or affect the pH. The benzoate ion (C\u2086H\u2085COO\u207b), however, is the conjugate base of a weak acid. It will react with water in a hydrolysis reaction, accepting a proton from water to form benzoic acid and hydroxide ions (OH\u207b).<\/p>\n\n\n\n

C\u2086H\u2085COO\u207b(aq) + H\u2082O(l) \u21cc C\u2086H\u2085COOH(aq) + OH\u207b(aq)<\/p>\n\n\n\n

The production of OH\u207b ions makes the solution basic. The equilibrium for this reaction is described by the base dissociation constant, K\u2091. The value of K\u2091 can be1.59 \u00d7 10\u207b\u00b9\u2070<\/p>\n\n\n\n

3. Equilibrium Calculation<\/strong>
We can now set up an ICE (Initial, Change, Equilibrium) table to determine the equilibrium concentration of OH\u207b. Let ‘x’ be the concentration of OH\u207b produced.<\/p>\n\n\n\n

<\/td>C\u2086H\u2085COO\u207b<\/td>C\u2086H\u2085COOH<\/td>OH\u207b<\/td><\/tr>
Initial (M)<\/strong><\/td>0.12<\/td>0<\/td>0<\/td><\/tr>
Change (M)<\/strong><\/td>-x<\/td>+x<\/td>+x<\/td><\/tr>
Equilibrium (M)<\/strong><\/td>0.12 – x<\/td>x<\/td>x<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

The K\u2091 expression is:
K\u2091 = [C\u2086H\u2085COOH][OH\u207b] \/ [C\u2086H\u2085COO\u207b]
1.59 \u00d7 10\u207b\u00b9\u2070 = (x)(x) \/ (0.12 – x)<\/p>\n\n\n\n

Since K\u2091 is very small, we can assume that x is negligible compared to 0.12 M (i.e., 0.12 – x \u2248 0.12).
1.59 \u00d7 10\u207b\u00b9\u2070 \u2248 x\u00b2 \/ 0.12
x\u00b2 = (1.59 \u00d7 10\u207b\u00b9\u2070)(0.12) = 1.908 \u00d7 10\u207b\u00b9\u00b9
x = \u221a(1.908 \u00d7 10\u207b\u00b9\u00b9) = 4.37 \u00d7 10\u207b\u2076 M<\/p>\n\n\n\n

So, [OH\u207b] = 4.37 \u00d7 10\u207b\u2076 M.<\/p>\n\n\n\n

4. Calculate pOH and pH<\/strong>
Now we can calculate the pOH and then the pH:
pOH = -log[OH\u207b] = -log(4.37 \u00d7 10\u207b\u2076) = 5.36
pH = 14.00 – pOH = 14.00 – 5.36 = 8.64<\/p>\n\n\n\n

The final pH of the 0.12 M sodium calculated from the acid dissociation constant (K\u2090) of its conjugate acid, benzoic acid, using the ion product of water (K\u2091 = 1.0 \u00d7 10\u207b\u00b9\u2074).<\/p>\n\n\n\n

K\u2091 = K\u2091 \/ K\u2090 = (1.0 \u00d7 10\u207b\u00b9\u2074) \/ (6.3 \u00d7 10\u207b\u2075) = 1.59 \u00d7 10\u207b\u00b9\u2070<\/p>\n\n\n\n

To find the hydroxide ion concentration, [OH\u207b], an ICE table is used, where ‘x’ represents the change in concentration at equilibrium.<\/p>\n\n\n\n

<\/td>C\u2086H\u2085COO\u207b<\/td>C\u2086H\u2085COOH<\/td>OH\u207b<\/td><\/tr>
Initial<\/strong><\/td>0.12 M<\/td>0<\/td>~0<\/td><\/tr>
Change<\/strong><\/td>-x<\/td>+x<\/td>+x<\/td><\/tr>
Equilibrium<\/strong><\/td>0.12 – x<\/td>x<\/td>x<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

The K\u2091 expression is:
K\u2091 = [C\u2086H\u2085COOH][OH\u207b] \/ [C\u2086H\u2085COO\u207b] = (x)(x) \/ (0.12 – x)<\/p>\n\n\n\n

Since K\u2091 is benzoate solution, rounded to three significant figures, is 8.64. very small, we can assume that x is negligible compared to the initial concentration of 0.12 M. Thus, (0.12 – x) \u2248 0.12.<\/p>\n\n\n\n

1.59 \u00d7 10\u207b\u00b9\u2070 = x\u00b2 \/ 0.12
x\u00b2 = (1.59 \u00d7 10\u207b\u00b9\u2070)(0.12) = 1.908 \u00d7 10\u207b\u00b9\u00b9
x = \u221a(1.908 \u00d7 10\u207b\u00b9\u00b9) = 4.37 \u00d7 10\u207b\u2076 M<\/p>\n\n\n\n

So, [OH\u207b] = 4.37 \u00d7 10\u207b\u2076 M.<\/p>\n\n\n\n

Next, the pOH is calculated:
pOH = -log[OH\u207b] = -log(4.37 \u00d7 10\u207b\u2076) = 5.36<\/p>\n\n\n\n

Finally, the pH is found using the relationship pH + pOH = 14.00:
pH = 14.00 – pOH = 14.00 – 5.36 = 8.64<\/p>\n\n\n\n

The pH of the 0.12thumb_upthumb_down<\/p>\n\n\n\n

\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

Sodium benzoateis a product of sodium hydroxide and benzoic acid(). What is the pH of 0.12 M Sodium benzoate? Please enter 3 significant figures (Sample problem 18.12 and follow-up problems) The Correct Answer and Explanation is: Answer: 8.64 Explanation: Sodium benzoate (C\u2086H\u2085COONa) is the salt formed from a strong base (sodium hydroxide, NaOH) and a weak […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-233026","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233026","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=233026"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233026\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=233026"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=233026"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=233026"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}