# barium hydroxide and hydrochloric acid

&= \pu{6e-4 mol} Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. HOWEVER, the book is saying $\mathrm{pH} = 12.55$. Our channel. I'm so lost when it comes to writing balanced and net ionic equations. Barium hydroxide react with hydrogen chloride. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. &= \pu{3.06e-3 mol} &= c(\ce{HCl})\cdot V(\ce{HCl}) \\ Ask Question Asked 2 years, 7 months ago. Copper and hydrochloric acid. This is an ionic substance so in water it breaks apart into ions H^+ + Cl_- Potassium Hydroxide is also an ionic substance it also breaks apart in water into ions K^+ + OH^- So the complete ionic equation for the reaction is H^+ + Cl^- + K^+ + OH^- = K^+ + Cl^- + H_2O The Hydrogen ion and the Hydroxide ions combine to form water. Write a net ionic equation for the reaction that occurs when aqueous solutions of barium hydroxide and hydrochloric acid are combined. calcium hydroxide + carbon dioxide = calcium carbonate + water; sulfur + ozone = sulfur dioxide Examples of the chemical equations reagents (a complete equation will be suggested): H 2 SO 4 + K 4 Fe(CN) 6 + KMnO 4; Ca(OH) 2 + H 3 PO 4; Na 2 S 2 O 3 + I 2; C … Calculating the pH upon titrating barium hydroxide and hydrochloric acid. Titration of barium(II) hydroxide by hydrochloric acid. In the nutshell, we are dealing with a typical neutralization reaction: and note that $\ce{BaCl2}$ as a salt of a strong acid and a strong base won't undergo hydrolysis and, as a consequence, won't have a noticeable impact on $\mathrm{pH}$. Chemical equation of reaction of BaO, 2HCl, BaCl2, H2O. . Looking at solution of excess $\ce{Ba(OH)2}$, Since $\ce{Ba(OH)2}$ : $\ce{2OH-}$, moles of $\ce{OH-} = 3.3 \times 10^{-4} \times 2 = 6.6 \times 10^{-4}$ moles of $\ce{OH-}$, So, concentration $= \frac{6.6 \times 10^{-4}}{\frac{25+45}{1000}} = \frac{6.6 \times 10^{-4}}{0.07} = 0.009..$, $\mathrm{pOH} = -\log \ce{[OH]} = -\log[0.009] = 2.026$, Since $\mathrm{pH} + \mathrm{pOH} = 14$, $\mathrm{pH} = 14 - \mathrm{pOH}$ &= 2\cdot\pu{3.4e-2 mol L-1}\cdot\pu{4.5e-2 L} \\ Generic word for firearms with long barrels. story about man trapped in dream. Help please! When barium hydroxide is titrated with hydrochloric acid, two molecules of hydrochloric acid combine with one molecule of barium hydroxide to produce one molecule of barium chloride and two molecules of water. 1 Answer to Hydrochloric acid, HCL, is reacted with barium hydroxide, Ba(OH)2. So $\mathrm{pH} = 14 - 2.026 = 11.97$. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. n_0(\ce{H3O+}) &= n(\ce{HCl}) \\ To learn more, see our tips on writing great answers. 2HCl(aq) + Ba(OH)2(aq) --> 2H2O(l) + BaCl2(aq) suppose a beaker contains 55 mL of 0.091 M Ba(OH)2. &= \pu{2.46e-3 mol} Reaction Equation and Ionic Equations: The reaction between barium hydroxide and hydrochloric acid gives barium chloride and water. I do have some questions. ${1.53} \times 10^{-3} - 1.2 \times 10^{-3} = 3.3 \times 10^{-4}$, this is the excess moles of $\ce{Ba(OH)2}$. ... Byproducts of the neutralisation of hydrochloric acid with sodium hydroxide. rev 2020.11.24.38066, The best answers are voted up and rise to the top, Chemistry Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, $\ce{Ba(OH)2} = \pu{0.034M} \times \frac{45}{1000} = 1.53 \times 10^{-3}$, $\ce{HCl} = 0.024 \times \frac{25}{1000} = 6 \times 10^{-4}$, ${1.53} \times 10^{-3} - 1.2 \times 10^{-3} = 3.3 \times 10^{-4}$, $\ce{OH-} = 3.3 \times 10^{-4} \times 2 = 6.6 \times 10^{-4}$, $= \frac{6.6 \times 10^{-4}}{\frac{25+45}{1000}} = \frac{6.6 \times 10^{-4}}{0.07} = 0.009..$, >"Moles of $\ce{HCl} = 0.024 \times \frac{25}{1000} = 6 \times 10^{-4}$ but $2\times6\times10^{-4}$ since it’s $\ce{2HCl}$ which is $1.2 \times 10^{-3}$".

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