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Question Number 30427 by rahul 19 last updated on 22/Feb/18

Commented by Tinkutara last updated on 22/Feb/18

Is the answer yes?

Commented by rahul 19 last updated on 22/Feb/18

$$\mathrm{actually},\mathrm{i}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{correct}\mathrm{ans}.$$$$\mathrm{it}\mathrm{came}\mathrm{in}\mathrm{my}\mathrm{final}\mathrm{exam}\mathrm{held}\mathrm{today}.$$$$\mathrm{can}\mathrm{u}\mathrm{show}\mathrm{the}\mathrm{working}?$$

Commented by Tinkutara last updated on 22/Feb/18

Commented by Tinkutara last updated on 23/Feb/18

$$\left[{Ba}^{2+}\right]=\frac{1.75mmol}{60mL}=\frac{7}{240}M$$$$\because Newvolume=25+35=60mL$$$$Wedonow\left[{Ba}^{2+}\right]{\left[F^{-}\right]}^{2}tocalculate$$$$thesolubilityproductof{BaF}_2.$$

Commented by rahul 19 last updated on 22/Feb/18

$$\mathrm{how}\left[{\mathrm{Ba}}^{2+}\right]=7/240?$$$$\mathrm{and}\mathrm{also}\mathrm{why}{\left[{\mathrm{F}}^{-}\right]}^2,\mathrm{acc}.\mathrm{to}\mathrm{balanced}\mathrm{eq}.$$$$\mathrm{it}\mathrm{should}\mathrm{be}\left[{\mathrm{F}}^{-}\right].$$

Commented by rahul 19 last updated on 23/Feb/18

$$\mathrm{ok},\mathrm{thanks}!$$

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