You have 100.0 mL of a solution of hydrochloric acid that has a pH of 3.00. You add 100.0 mL of water to this solution. What is the resulting pH of the solution?<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To determine the resulting pH after diluting a hydrochloric acid (HCl) solution with water, we need to start by understanding the concentration of the original solution.<\/p>\n\n\n\n Step 1: Calculate the concentration of HCl in the original solution.<\/strong><\/p>\n\n\n\n The pH of a solution is related to the concentration of hydrogen ions ([H^+]) through the equation:<\/p>\n\n\n\n [ Given that the pH is 3.00, we can calculate the concentration of hydrogen ions:<\/p>\n\n\n\n [ Since HCl is a strong acid, it dissociates completely in solution. Therefore, the concentration of HCl in the original solution is also 0.001 M.<\/p>\n\n\n\n Step 2: Calculate the total volume after dilution.<\/strong><\/p>\n\n\n\n When you add 100.0 mL of water to 100.0 mL of the HCl solution, the total volume becomes:<\/p>\n\n\n\n [ Step 3: Calculate the new concentration of HCl after dilution.<\/strong><\/p>\n\n\n\n The moles of HCl remain constant before and after dilution. Initially, we have:<\/p>\n\n\n\n [ After dilution, the concentration of HCl in the total volume of 200.0 mL is:<\/p>\n\n\n\n [ Step 4: Calculate the new pH.<\/strong><\/p>\n\n\n\n Now we can find the new pH using the new concentration of hydrogen ions:<\/p>\n\n\n\n [ Conclusion:<\/strong><\/p>\n\n\n\n The resulting pH of the solution after adding 100.0 mL of water to the original 100.0 mL solution of hydrochloric acid with a pH of 3.00 is approximately 3.30. This demonstrates that dilution of an acid leads to an increase in pH, reflecting a decrease in hydrogen ion concentration.<\/p>\n","protected":false},"excerpt":{"rendered":" You have 100.0 mL of a solution of hydrochloric acid that has a pH of 3.00. You add 100.0 mL of water to this solution. What is the resulting pH of the solution? The Correct Answer and Explanation is : To determine the resulting pH after diluting a hydrochloric acid (HCl) solution with water, we […]<\/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-153017","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/153017","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=153017"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/153017\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=153017"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=153017"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=153017"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\\text{pH} = -\\log[H^+]
]<\/p>\n\n\n\n
[H^+] = 10^{-\\text{pH}} = 10^{-3} \\, \\text{M} = 0.001 \\, \\text{M}
]<\/p>\n\n\n\n
\\text{Total Volume} = 100.0 \\, \\text{mL} + 100.0 \\, \\text{mL} = 200.0 \\, \\text{mL}
]<\/p>\n\n\n\n
\\text{Moles of HCl} = \\text{Concentration} \\times \\text{Volume} = 0.001 \\, \\text{M} \\times 0.1 \\, \\text{L} = 0.0001 \\, \\text{moles}
]<\/p>\n\n\n\n
\\text{New Concentration} = \\frac{\\text{Moles of HCl}}{\\text{Total Volume}} = \\frac{0.0001 \\, \\text{moles}}{0.2 \\, \\text{L}} = 0.0005 \\, \\text{M}
]<\/p>\n\n\n\n
\\text{pH} = -\\log[H^+] = -\\log(0.0005) \\approx 3.30
]<\/p>\n\n\n\n