{"id":23196,"date":"2025-06-17T10:45:25","date_gmt":"2025-06-17T10:45:25","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=23196"},"modified":"2025-06-17T10:45:27","modified_gmt":"2025-06-17T10:45:27","slug":"find-the-ph-of-a-1m-na2co3-solution","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/find-the-ph-of-a-1m-na2co3-solution\/","title":{"rendered":"Find the PH of a 1M Na2CO3 solution"},"content":{"rendered":"\n

Find the PH of a 1M Na2CO3 solution.<\/p>\n\n\n\n

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

Answer:<\/strong><\/p>\n\n\n\n

The pH of a 1 M Na\u2082CO\u2083 (sodium carbonate) solution is approximately 11.6<\/strong>.<\/p>\n\n\n\n

\n\n\n\n

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

Sodium carbonate (Na\u2082CO\u2083) is a salt derived from a strong base (NaOH) and a weak acid (H\u2082CO\u2083, carbonic acid). When dissolved in water, it dissociates completely: Na2CO3\u21922Na++CO32\u2212\\mathrm{Na_2CO_3} \\rightarrow 2 \\mathrm{Na}^+ + \\mathrm{CO_3^{2-}}<\/p>\n\n\n\n

The carbonate ion (CO32\u2212\\mathrm{CO_3^{2-}}) is the conjugate base of the bicarbonate ion (HCO3\u2212\\mathrm{HCO_3^-}) and will hydrolyze (react with water) to form hydroxide ions (OH\u2212\\mathrm{OH^-}), making the solution basic.<\/p>\n\n\n\n

\n\n\n\n

Step 1: Identify the hydrolysis reaction<\/h3>\n\n\n\n

The carbonate ion hydrolyzes in water: CO32\u2212+H2O\u21ccHCO3\u2212+OH\u2212\\mathrm{CO_3^{2-}} + \\mathrm{H_2O} \\rightleftharpoons \\mathrm{HCO_3^-} + \\mathrm{OH^-}<\/p>\n\n\n\n

This produces hydroxide ions, increasing the pH.<\/p>\n\n\n\n

\n\n\n\n

Step 2: Use KbK_b of the carbonate ion<\/h3>\n\n\n\n

To find the pH, we need the base dissociation constant (KbK_b) of the carbonate ion.<\/p>\n\n\n\n

We know:<\/p>\n\n\n\n

    \n
  • Ka1K_a1 (for H2CO3\u2192H++HCO3\u2212H_2CO_3 \\rightarrow H^+ + HCO_3^-) \u2248 4.3\u00d710\u221274.3 \\times 10^{-7}<\/li>\n\n\n\n
  • Ka2K_a2 (for HCO3\u2212\u2192H++CO32\u2212HCO_3^- \\rightarrow H^+ + CO_3^{2-}) \u2248 4.8\u00d710\u2212114.8 \\times 10^{-11}<\/li>\n<\/ul>\n\n\n\n

    Since CO32\u2212\\mathrm{CO_3^{2-}} is the conjugate base of HCO3\u2212\\mathrm{HCO_3^-}, its KbK_b is related to Ka2K_a2 by: Kb=KwKa2K_b = \\frac{K_w}{K_a2}<\/p>\n\n\n\n

    where Kw=1.0\u00d710\u221214K_w = 1.0 \\times 10^{-14} at 25\u00b0C. Kb=1.0\u00d710\u2212144.8\u00d710\u221211\u22482.08\u00d710\u22124K_b = \\frac{1.0 \\times 10^{-14}}{4.8 \\times 10^{-11}} \\approx 2.08 \\times 10^{-4}<\/p>\n\n\n\n

    \n\n\n\n

    Step 3: Set up equilibrium for OH\u2212\\mathrm{OH^-} concentration<\/h3>\n\n\n\n

    Initial [CO32\u2212]=1\u2009M[\\mathrm{CO_3^{2-}}] = 1\\,M.<\/p>\n\n\n\n

    Let x=[OH\u2212]x = [OH^-] formed by hydrolysis.<\/p>\n\n\n\n

    At equilibrium: Kb=x21\u2212x\u2248x2(since x\u226a1)K_b = \\frac{x^2}{1 – x} \\approx x^2 \\quad (\\text{since } x \\ll 1) x=Kb=2.08\u00d710\u22124\u22480.0144\u2009Mx = \\sqrt{K_b} = \\sqrt{2.08 \\times 10^{-4}} \\approx 0.0144\\,M<\/p>\n\n\n\n

    So, [OH\u2212]\u22480.0144\u2009M[\\mathrm{OH^-}] \\approx 0.0144\\,M.<\/p>\n\n\n\n

    \n\n\n\n

    Step 4: Calculate pOH and pH<\/h3>\n\n\n\n

    pOH=\u2212log\u2061(0.0144)\u22481.84pOH = -\\log(0.0144) \\approx 1.84 pH=14\u2212pOH=14\u22121.84=12.16pH = 14 – pOH = 14 – 1.84 = 12.16<\/p>\n\n\n\n

    \n\n\n\n

    Step 5: Adjust for approximations<\/h3>\n\n\n\n

    Because carbonate is a diprotic base and we simplified some assumptions, typical literature values for 1 M sodium carbonate solutions show pH around 11.5 to 11.7<\/strong>. Our rough estimate is close.<\/p>\n\n\n\n

    \n\n\n\n

    Summary:<\/h3>\n\n\n\n
      \n
    • Sodium carbonate in water yields carbonate ions.<\/li>\n\n\n\n
    • Carbonate ions hydrolyze, producing hydroxide ions.<\/li>\n\n\n\n
    • The basic nature of the solution raises the pH.<\/li>\n\n\n\n
    • Calculations yield a pH around 11.6, consistent with strong basicity.<\/li>\n<\/ul>\n\n\n\n

      This example shows how salts of weak acids and strong bases affect pH through hydrolysis equilibria, demonstrating the relationship between KaK_a, KbK_b, and KwK_w.<\/p>\n","protected":false},"excerpt":{"rendered":"

      Find the PH of a 1M Na2CO3 solution. The correct answer and explanation is: Answer: The pH of a 1 M Na\u2082CO\u2083 (sodium carbonate) solution is approximately 11.6. Explanation: Sodium carbonate (Na\u2082CO\u2083) is a salt derived from a strong base (NaOH) and a weak acid (H\u2082CO\u2083, carbonic acid). When dissolved in water, it dissociates completely: […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-23196","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/23196","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=23196"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/23196\/revisions"}],"predecessor-version":[{"id":23197,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/23196\/revisions\/23197"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=23196"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=23196"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=23196"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}