i. When aqueous solution of lead nitrate Pb(NO3)2 and potassium chloride KCl are mixed, a precipitate of lead chloride is formed. Write a balanced net ionic equation for this reaction.<\/p>\n\n\n\n
ii. 13.0 mL of hydrochloric acid (HCI) required to neutralize 36.11 ml of 0.045 M sodium hydroxide (NaOH). What
is the molarity of hydrochloric acid?<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n When aqueous solutions of lead nitrate (( \\text{Pb(NO}_3)_2 )) and potassium chloride (( \\text{KCl} )) are mixed, a double displacement reaction occurs, forming lead chloride (( \\text{PbCl}_2 )) as a precipitate and potassium nitrate (( \\text{KNO}_3 )) in solution. The balanced molecular equation for the reaction is:<\/p>\n\n\n\n [ The ionic equation shows all the ions present in the reaction:<\/p>\n\n\n\n [ Finally, the net ionic equation eliminates the spectator ions (in this case, ( \\text{K}^+ ) and ( \\text{NO}_3^- )) to show only the ions involved in the formation of the precipitate:<\/p>\n\n\n\n [ This net ionic equation represents the reaction where lead ions react with chloride ions to form solid lead chloride.<\/p>\n\n\n\n The balanced chemical equation for the neutralization of hydrochloric acid (( \\text{HCl} )) with sodium hydroxide (( \\text{NaOH} )) is:<\/p>\n\n\n\n [ From the equation, we see that hydrochloric acid reacts with sodium hydroxide in a 1:1 molar ratio. Given:<\/p>\n\n\n\n We can calculate the number of moles of sodium hydroxide:<\/p>\n\n\n\n [ Since the reaction is in a 1:1 molar ratio, the moles of ( \\text{HCl} ) required to neutralize the sodium hydroxide are also 0.001625 mol. Using the formula for molarity:<\/p>\n\n\n\n [ Thus, the molarity of hydrochloric acid is 0.125 M<\/strong>.<\/p>\n\n\n\n The key steps to solving this problem involve recognizing the stoichiometric relationship between ( \\text{HCl} ) and ( \\text{NaOH} ), which is 1:1 in the neutralization reaction. Using the given volume and molarity of sodium hydroxide, we can calculate the moles of sodium hydroxide present. From there, we know that an equal number of moles of hydrochloric acid is needed for neutralization. Finally, using the volume of hydrochloric acid and the calculated moles, we can find its molarity.<\/p>\n","protected":false},"excerpt":{"rendered":" i. When aqueous solution of lead nitrate Pb(NO3)2 and potassium chloride KCl are mixed, a precipitate of lead chloride is formed. Write a balanced net ionic equation for this reaction. ii. 13.0 mL of hydrochloric acid (HCI) required to neutralize 36.11 ml of 0.045 M sodium hydroxide (NaOH). Whatis the molarity of hydrochloric acid? The […]<\/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-189599","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189599","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=189599"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189599\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=189599"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=189599"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=189599"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Part i: Balanced Net Ionic Equation for the Reaction Between Lead Nitrate and Potassium Chloride<\/h3>\n\n\n\n
\\text{Pb(NO}_3)_2 (aq) + 2 \\text{KCl} (aq) \\rightarrow \\text{PbCl}_2 (s) + 2 \\text{KNO}_3 (aq)
]<\/p>\n\n\n\n
\\text{Pb}^{2+} (aq) + 2 \\text{NO}_3^- (aq) + 2 \\text{K}^+ (aq) + 2 \\text{Cl}^- (aq) \\rightarrow \\text{PbCl}_2 (s) + 2 \\text{K}^+ (aq) + 2 \\text{NO}_3^- (aq)
]<\/p>\n\n\n\n
\\text{Pb}^{2+} (aq) + 2 \\text{Cl}^- (aq) \\rightarrow \\text{PbCl}_2 (s)
]<\/p>\n\n\n\nPart ii: Calculating the Molarity of Hydrochloric Acid<\/h3>\n\n\n\n
\\text{HCl} (aq) + \\text{NaOH} (aq) \\rightarrow \\text{NaCl} (aq) + \\text{H}_2\\text{O} (l)
]<\/p>\n\n\n\n\n
\\text{moles of NaOH} = \\text{molarity of NaOH} \\times \\text{volume of NaOH} = 0.045 \\, \\text{M} \\times 0.03611 \\, \\text{L} = 0.001625 \\, \\text{mol}
]<\/p>\n\n\n\n
\\text{Molarity of HCl} = \\frac{\\text{moles of HCl}}{\\text{volume of HCl in liters}} = \\frac{0.001625 \\, \\text{mol}}{0.0130 \\, \\text{L}} = 0.125 \\, \\text{M}
]<\/p>\n\n\n\nExplanation:<\/h3>\n\n\n\n