{"id":22546,"date":"2025-06-16T19:47:58","date_gmt":"2025-06-16T19:47:58","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=22546"},"modified":"2025-06-16T19:48:00","modified_gmt":"2025-06-16T19:48:00","slug":"50-ml-of-0-250-m-ethylenediamine-solution-was-mixed-with-12-5ml-of-1-00-m-hcl","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/50-ml-of-0-250-m-ethylenediamine-solution-was-mixed-with-12-5ml-of-1-00-m-hcl\/","title":{"rendered":"50 mL of 0.250 M ethylenediamine solution was mixed with 12.5mL of 1.00 M HCl"},"content":{"rendered":"\n<p>50 mL of 0.250 M ethylenediamine solution was mixed with 12.5mL of 1.00 M HCl. WHat is the pH of the solution<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\"><strong>The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n<p>Let&#8217;s analyze the problem step-by-step:<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Given:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Volume of ethylenediamine (EN) solution, VEN=50.0\u2009mLV_{EN} = 50.0 \\, mL<\/li>\n\n\n\n<li>Concentration of EN, CEN=0.250\u2009MC_{EN} = 0.250 \\, M<\/li>\n\n\n\n<li>Volume of HCl solution, VHCl=12.5\u2009mLV_{HCl} = 12.5 \\, mL<\/li>\n\n\n\n<li>Concentration of HCl, CHCl=1.00\u2009MC_{HCl} = 1.00 \\, M<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Goal:<\/h3>\n\n\n\n<p>Find the <strong>pH<\/strong> of the final solution after mixing.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Understand the chemistry<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Ethylenediamine (EN) is a <strong>weak base<\/strong> with two amine groups NH2\\mathrm{NH_2}. It can accept protons from HCl.<\/li>\n\n\n\n<li>HCl is a <strong>strong acid<\/strong>, so it will fully dissociate into H+H^+ ions.<\/li>\n\n\n\n<li>When EN and HCl are mixed, HCl will protonate EN, forming the protonated ethylenediamine and Cl\u2212Cl^-.<\/li>\n\n\n\n<li>This is essentially a <strong>buffer system<\/strong> made from the weak base (EN) and its conjugate acid (protonated EN).<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Calculate moles of EN and HCl<\/h3>\n\n\n\n<p>nEN=CEN\u00d7VEN=0.250\u2009M\u00d70.0500\u2009L=0.0125\u2009moln_{EN} = C_{EN} \\times V_{EN} = 0.250 \\, M \\times 0.0500 \\, L = 0.0125 \\, mol nHCl=CHCl\u00d7VHCl=1.00\u2009M\u00d70.0125\u2009L=0.0125\u2009moln_{HCl} = C_{HCl} \\times V_{HCl} = 1.00 \\, M \\times 0.0125 \\, L = 0.0125 \\, mol<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Determine the limiting reagent and the reaction<\/h3>\n\n\n\n<p>Since nEN=nHCl=0.0125\u2009moln_{EN} = n_{HCl} = 0.0125 \\, mol, they are in a <strong>1:1 molar ratio<\/strong>.<\/p>\n\n\n\n<p>The reaction: EN+HCl\u2192ENH++Cl\u2212\\text{EN} + \\text{HCl} \\rightarrow \\text{ENH}^+ + \\text{Cl}^-<\/p>\n\n\n\n<p>Because the moles of EN and HCl are equal, all the ethylenediamine is protonated, forming 0.01250.0125 mol of its conjugate acid, and no free EN remains.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Calculate concentration of species in the total volume<\/h3>\n\n\n\n<p>Total volume: Vtotal=50.0\u2009mL+12.5\u2009mL=62.5\u2009mL=0.0625\u2009LV_{total} = 50.0 \\, mL + 12.5 \\, mL = 62.5 \\, mL = 0.0625 \\, L<\/p>\n\n\n\n<p>Concentration of protonated EN (the conjugate acid): CENH+=0.0125\u2009mol0.0625\u2009L=0.20\u2009MC_{ENH^+} = \\frac{0.0125 \\, mol}{0.0625 \\, L} = 0.20 \\, M<\/p>\n\n\n\n<p>No free EN remains.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 5: Determine pH<\/h3>\n\n\n\n<p>Since all EN is converted to the conjugate acid ENH+ENH^+, the solution is <strong>acidic<\/strong>, dominated by the protonated species.<\/p>\n\n\n\n<p>To find the pH, we need the <strong>acid dissociation constant KaK_a<\/strong> of the conjugate acid.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Ethylenediamine has Kb\u22487.9\u00d710\u22124K_b \\approx 7.9 \\times 10^{-4}.<\/li>\n\n\n\n<li>Use Ka\u00d7Kb=Kw=1.0\u00d710\u221214K_a \\times K_b = K_w = 1.0 \\times 10^{-14} at 25\u00b0C.<\/li>\n<\/ul>\n\n\n\n<p>Calculate KaK_a: Ka=KwKb=1.0\u00d710\u2212147.9\u00d710\u22124\u22481.27\u00d710\u221211K_a = \\frac{K_w}{K_b} = \\frac{1.0 \\times 10^{-14}}{7.9 \\times 10^{-4}} \\approx 1.27 \\times 10^{-11}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 6: Calculate pH using KaK_a<\/h3>\n\n\n\n<p>Using the KaK_a for ENH+ENH^+: Ka=[H+][EN][ENH+]K_a = \\frac{[H^+][EN]}{[ENH^+]}<\/p>\n\n\n\n<p>Initially, we only have ENH+ENH^+ at 0.20 M. ENEN and H+H^+ come from dissociation of ENH+ENH^+.<\/p>\n\n\n\n<p>Let x=[H+]x = [H^+]: Ka=x20.20\u2212x\u2248x20.20K_a = \\frac{x^2}{0.20 &#8211; x} \\approx \\frac{x^2}{0.20}<\/p>\n\n\n\n<p>(Assuming xx is very small compared to 0.20)<\/p>\n\n\n\n<p>Solve for xx: x2=Ka\u00d70.20=1.27\u00d710\u221211\u00d70.20=2.54\u00d710\u221212x^2 = K_a \\times 0.20 = 1.27 \\times 10^{-11} \\times 0.20 = 2.54 \\times 10^{-12} x=2.54\u00d710\u221212=1.59\u00d710\u22126\u2009Mx = \\sqrt{2.54 \\times 10^{-12}} = 1.59 \\times 10^{-6} \\, M<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 7: Calculate pH<\/h3>\n\n\n\n<p>pH=\u2212log\u2061[H+]=\u2212log\u2061(1.59\u00d710\u22126)\u22485.80pH = -\\log[H^+] = -\\log(1.59 \\times 10^{-6}) \\approx 5.80<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h1 class=\"wp-block-heading\"><strong>Final answer:<\/strong><\/h1>\n\n\n\n<p><strong>pH \u2248 5.80<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation summary:<\/h3>\n\n\n\n<p>You start with ethylenediamine, a weak base, and add a strong acid (HCl). Because the moles of acid and base are equal, the acid completely protonates the base, converting all EN to its conjugate acid ENH+ENH^+. The solution behaves like a weak acid solution, so the pH is calculated using the acid dissociation constant derived from KbK_b of EN.<\/p>\n\n\n\n<p>The calculated pH (~5.80) is acidic, but not strongly so, because ethylenediamine is a weak base and its conjugate acid is a weak acid. This explains the mildly acidic pH after mixing.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>50 mL of 0.250 M ethylenediamine solution was mixed with 12.5mL of 1.00 M HCl. WHat is the pH of the solution The correct answer and explanation is: Let&#8217;s analyze the problem step-by-step: Given: Goal: Find the pH of the final solution after mixing. Step 1: Understand the chemistry Step 2: Calculate moles of EN [&hellip;]<\/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-22546","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/22546","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=22546"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/22546\/revisions"}],"predecessor-version":[{"id":22547,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/22546\/revisions\/22547"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=22546"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=22546"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=22546"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}