<u>Answer:</u>
<em>To raise the pH of the solution to 3.10 we have to add 2.34 L of water.</em>
<u>Explanation:</u>
<em>Given that the pH of the solution of HCl in water is 2.5.</em> Here the solution’s pH is changing from 2.5 to 3.10 which means the acidic nature of the solution is decreasing here on dilution. ![[H^+]](https://tex.z-dn.net/?f=%5BH%5E%2B%5D)
ions contribute to a solution’s acidic nature and ![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D)
contribute to a solution’s basic nature.
The equation connecting the concentration of and pH of a solution is pH= ![-log[H^+]](https://tex.z-dn.net/?f=-log%5BH%5E%2B%5D)
<em>![[H^+]= 10^(^-^p^H^)](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%2010%5E%28%5E-%5Ep%5EH%5E%29)
![[H^+]= 10^(^-^2^.^5^)=0.00316](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%2010%5E%28%5E-%5E2%5E.%5E5%5E%29%3D0.00316)
</em>
<em>When the pH is ![3.1 [H^+ ]= 10^(^-^3^.^1^)=0.000794](https://tex.z-dn.net/?f=3.1%20%5BH%5E%2B%20%5D%3D%2010%5E%28%5E-%5E3%5E.%5E1%5E%29%3D0.000794)
</em>
<em>On dilution the concentration of a solution decreases and volume increases.</em>
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<em>Volume of water to be added 
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Balanced equation:
<span>2 NO + 5 H2 ------> 2 NH3 + 2 H2O
</span>
<span>2 moles NO react with 5 moles H2 to produce 2 moles NH3
</span>
<span>Molar mass of NO = 30.00 g/mol </span>
<span>86.3g NO = 86.3/30.00 = 2.877 moles of NO </span>
<span>This will require: 2.877*5 / 2 = 7.192 moles of H2 </span>
<span>Molar mass of H2 = 2 g/mol </span>
<span>25.6g H2 = 25.6/2 = 12.7 mol H2. </span>
<span>You have excess H2 means the NO is limiting </span>
<span>From the balanced equation: </span>
<span>2 moles of NO will produce 2 moles of NH3 </span>
<span>2.877 moles of NO will produce 2.877 moles of NH3 </span>
<span>Molar mass NH3 = 17g/mol </span>
<span>Mass NH3 produced = 2.877 * 17 = 48.91g
Hence the yield is = 48.91 g ~ 49 g</span>
