Q 7 please 1. The pH of a solution is 4.80. What is the concentration of hydroxide ions in this solution? a.4.2 x 10 M b. 1.6 x 105 M c. 3.6 x 10-12 M d. 6.3 x 10-10 M 2 the pH of a 0.50 M solution of NaNO\u00e2\u201a\u201a. Ka for HNO\u00e2\u201a\u201a is 4\u00c3\u201410-5 a 12.1 b. 5.48 c. 1.82 c. 8.90 d 8.52 3- What was the pH of the solution that result from titration of 25.0 ml of 0.5 M solution of weak base (Kb – 9.74\u00c3\u201410) wi 30 ml of 0.1 M hydrochloric acid, HCI.? a. 5.10 b. 4.92 d. 9.1 4-Which of the following combinations cannot produce a buffer solution? a. HNO2 and NaNO\u00e2\u201a\u201a b. HCN and NaCN c. HCIO4 and NaCIO4 d. NH3 and (NH4)2SO4 e. 2.0 x 10-8 M d. 5.5 e. NH3 and NH4Br 5- What is the pH at the equivalence point in the titration of 100.0 mL of 0.20 M ammonia (NH3) with 0.10 M hydrochloric acid (HCI)? Kb for NH31.8x 10-5 a. 4.6 b. 5.2 c. 7.0 e. 4.9 e.7.00 9. what will happened to the pH of buffer system when it diluted a. ApH=0 b. pH will increased c. pH will decrease 6- If 10.00 ml of 0.1M NaOH solution is needed to reach the end point of 10 ml HNO3. Then the concentration of HNO3 in ppm is: [M.wt. of HNO3= 63 g\/mol] e. 25.77 x 10\u00c2\u00b3 ppm a. 9.45\u00c3\u201410\u00c2\u00b3 ppm b.14.70 x10\u00c2\u00b3 ppm c. 6.30 \u00c3\u2014 10\u00c2\u00b3 ppm d. 18.77 \u00c3\u2014 10\u00c2\u00b9 ppm 7. A 0.6745-gram sample of KHP reacts with 41.75 mL of KOH solution for complete neutralization. What is the molarity of the KOH solution? (Molecular weight of KHP = 204 g\/mol. KHP has one acidic hydrogen.) a. 0.158 M c. 0.139 M b. 0.099 M d. 0.079 M e. 0.061 M 8. Which of the following combinations would give a pH =7.00 at the “equivalence point” (when equal moles of each have been added)? a. HCI + KF b. HCN + NaOH c. HF + HCI d. HCI + KOH 13. The molar analytical concentration of NO were added to 20.0 ml of 0. 82 M Fe (NO3)2 a.1.09M b. 2.53 M d. pH >7 e. pH = 7 10. the pH for Acidic buffer system with highest capacity if the Ka for the weak Acid 1.8 104 is a. 4.3 b.5.3 c.3.77 d.6.8 e. 8.3 e. 9.5 c.0.73 M 11.The ppm concentration of Cl-ion in 100 ml mixture solution of 0.01 M CaCl2 and 0.1 M HCI is (MW of Cl = 35.45 g\/mol and for Cl =35.45 g\/mol ) is :- b. 4254 ppm c. 355 ppm d. 9150ppm a. 243 ppm e. 850ppm from the primary- 12. The mass of AgNO3(s) needed to prepare 1.000 L of 0.0500 M AgNO3 (169.87 g\/mol) solution standard-grade solid is a. 18.548g b. 16.987 g c. 16.139 g d. 9.843g e. 10.36 g ions in the solution produced when 25.0 ml of Distilled water d. 0.11 M 14. Standard solution of EDTA (0.100 M) is being used to titrate 25.00 ml of (0.0100) M of Zn\u00c2\u00b2+ solution (buffered at pH =10, 04-0.36 and kMY = 3.2×10\u00c2\u00b96) the pZn after the addition of 25 ml EDTA solution is (at equivalent point) a. 8.68 b. 9.03 d. 9.79 c. 3.76 e. 0.13 M e. 6.89<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n A 0.6745-gram sample of KHP<\/strong> reacts with 41.75 mL of KOH solution<\/strong> for complete neutralization. The molarity of the KOH solution<\/strong> can be calculated using the molecular weight of KHP, which is 204 g\/mol<\/strong>. KHP reacts with KOH in a 1:1 molar ratio since KHP contains one acidic hydrogen.<\/p>\n\n\n\n d. 0.079 M<\/strong><\/p>\n\n\n\n The calculation revolves around stoichiometry and molarity. KHP (Potassium Hydrogen Phthalate) is a primary standard acid used in titrations due to its high purity and one acidic proton. The reaction with KOH is straightforward:<\/p>\n\n\n\n [ We first find the moles of KHP using its mass and molar mass, then calculate the molarity of KOH based on the volume of KOH used. The final molarity of 0.079 M fits perfectly with the neutralization reaction and the given data. This step-by-step method ensures accuracy, critical for quantitative chemical analysis.<\/p>\n\n\n\n <\/p>\n","protected":false},"excerpt":{"rendered":" Q 7 please 1. The pH of a solution is 4.80. What is the concentration of hydroxide ions in this solution? a.4.2 x 10 M b. 1.6 x 105 M c. 3.6 x 10-12 M d. 6.3 x 10-10 M 2 the pH of a 0.50 M solution of NaNO\u00e2\u201a\u201a. Ka for HNO\u00e2\u201a\u201a is 4\u00c3\u201410-5 […]<\/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-183319","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183319","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=183319"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183319\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=183319"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=183319"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=183319"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Question 7:<\/h3>\n\n\n\n
Step-by-step solution:<\/h4>\n\n\n\n
\n
[
\\text{Moles of KHP} = \\frac{\\text{Mass of KHP}}{\\text{Molar mass of KHP}} = \\frac{0.6745 \\, \\text{g}}{204 \\, \\text{g\/mol}} = 0.003307 \\, \\text{mol}
]<\/li>\n\n\n\n
Since the moles of KHP are equal to the moles of KOH due to the 1:1 stoichiometry:
[
\\text{Molarity of KOH} = \\frac{\\text{Moles of KOH}}{\\text{Volume of KOH in liters}}
]
Volume of KOH in liters:
[
41.75 \\, \\text{mL} = 0.04175 \\, \\text{L}
]
Substituting:
[
\\text{Molarity of KOH} = \\frac{0.003307}{0.04175} = 0.079 \\, \\text{M}
]<\/li>\n<\/ol>\n\n\n\nFinal Answer:<\/h4>\n\n\n\n
\n\n\n\nExplanation:<\/h3>\n\n\n\n
\\text{KHP} + \\text{KOH} \\to \\text{KP} + \\text{H}_2\\text{O}
]<\/p>\n\n\n\n