1f:I[80107,["/_next/static/chunks/235ad55241b6443f.js","/_next/static/chunks/8337ab1f037b6654.js","/_next/static/chunks/c9e933b5fd1a0afc.js","/_next/static/chunks/848a330faa2461ac.js","/_next/static/chunks/9f2942e368a262d2.js","/_next/static/chunks/92f9e11244990470.js","/_next/static/chunks/a8b867fbb1dfca1a.js","/_next/static/chunks/1100f0390a901fa7.js","/_next/static/chunks/bbe6bd9d3c7b0355.js","/_next/static/chunks/a82195c3e5de7dd4.js"],"default"] 20:I[76763,["/_next/static/chunks/235ad55241b6443f.js","/_next/static/chunks/8337ab1f037b6654.js","/_next/static/chunks/c9e933b5fd1a0afc.js","/_next/static/chunks/848a330faa2461ac.js","/_next/static/chunks/9f2942e368a262d2.js","/_next/static/chunks/92f9e11244990470.js","/_next/static/chunks/a8b867fbb1dfca1a.js","/_next/static/chunks/1100f0390a901fa7.js","/_next/static/chunks/bbe6bd9d3c7b0355.js","/_next/static/chunks/a82195c3e5de7dd4.js"],"default"] 1e:T446,{"@context":"https://schema.org","@graph":[{"@type":"BreadcrumbList","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https://www.divinejee.com"},{"@type":"ListItem","position":4,"name":"250 mL of 0050 M BaNO32 is mixed with 250 mL of 0020 M NaF K","item":"https://www.divinejee.com/question/250-ml-of-0050-m-bano32-is-mixed-with-250-ml-of-0020-m-naf-ksp-of-baf2-is-05-x-106-at-298-k-the-rat"}]},{"@type":"QAPage","mainEntity":{"@type":"Question","name":"Question 250 mL of 0050 M BaNO32 is mixed with 250 mL of 0020 M NaF K on Topic","text":"

$25.0 \\mathrm{~mL}$ of $0.050 ~\\mathrm{M} ~\\mathrm{Ba}\\left(\\mathrm{NO}_{3}\\right)_{2}$ is mixed with $25.0 \\mathrm{~mL}$ of $0.020 ~\\mathrm{M} ~\\m...","answerCount":1,"upvoteCount":0,"datePublished":"2026-01-16","author":{"@type":"Organization","name":"DivineJEE"},"acceptedAnswer":{"@type":"Answer","text":"Check the detailed solution on the page.","upvoteCount":0,"url":"https://www.divinejee.com/question/250-ml-of-0050-m-bano32-is-mixed-with-250-ml-of-0020-m-naf-ksp-of-baf2-is-05-x-106-at-298-k-the-rat"}}}]}21:T51e,Initial concentrations before mixing:

$$[\mathrm{Ba(NO_3)_2}] = 0.050\, \mathrm{M}$$

$$[\mathrm{NaF}] = 0.020\, \mathrm{M}$$

Volumes of the solutions:

$$V_{\mathrm{Ba(NO_3)_2}} = V_{\mathrm{NaF}} = 25.0\, \mathrm{mL}$$

After mixing, the total volume becomes:

$$V_{\mathrm{total}} = 25.0\, \mathrm{mL} + 25.0\, \mathrm{mL} = 50.0\, \mathrm{mL}$$

Now, we calculate the initial concentrations after mixing:

$$[\mathrm{Ba}^{2+}] = \frac{25.0\, \mathrm{mL} \times 0.050\, \mathrm{M}}{50.0\, \mathrm{mL}} = 0.025\, \mathrm{M}$$

$$[\mathrm{F}^{-}] = \frac{25.0\, \mathrm{mL} \times 0.020\, \mathrm{M}}{50.0\, \mathrm{mL}} = 0.010\, \mathrm{M}$$

Next, we calculate the reaction quotient (Q) for the precipitation of BaF₂:

$$Q = [\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^2$$

$$Q = (0.025\, \mathrm{M})(0.010\, \mathrm{M})^2 = 2.5 \times 10^{-6}$$

K_sp of BaF₂ is given as $$5 \times 10^{-7}$$.

Now, we find the ratio of $$[\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^2$$ to K_sp:

$$\text{Ratio} = \frac{(2.5 \times 10^{-6})}{(5 \times 10^{-7})} = 5$$

So, the correct ratio of $$[\mathrm{Ba}^{2+}][\mathrm{F}^{-}]^2$$ to K_sp is 5.