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anonymous

  • one year ago

How many grams of pure NaOH must be used to prepare 10.0 L of a solution that has a pH of 13?

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  1. JoannaBlackwelder
    • one year ago
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    pH=-log[H+]

  2. JoannaBlackwelder
    • one year ago
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    1.0x10^-14=[H+][OH-]

  3. JoannaBlackwelder
    • one year ago
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    M=mol/L

  4. JoannaBlackwelder
    • one year ago
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    Any ideas on how to use these formulas for this problem?

  5. Rushwr
    • one year ago
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    Hey , The 1st thing to pop when u read this question is that we can't do anything with pH since it is a base. So u have to find a way to get the pOH from pH for that we can use the following, \[14 =pH+pOH\] (since they haven't given us any detail about the temp and pressure, we will take it at stp so ) Now we found that pOH is equal to 14-13= 1

  6. Rushwr
    • one year ago
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    Now we can find the concentration of OH using \[pOH = -\log_{10}[OH ^{-}]\] So will be getting an answer around 0.1 moldm-3 So now we know the concentration of OH- and we also know the volume of of it which 10L Now using \[C * V= n\] find the no. of moles (n), which will be 1mol We know the no. of moles of OH- is equal to the no. of moles of NaOH Now using \[n=\frac{ m }{ M _{w} }\] find the mass (m). As we know the r.m.m. of NaOH which is 40gmol-1 So the mass of NaOH will be 1mol x 40gmol-1 which is equal to 40gmole again ! simple right? HOpe u got it

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