Answer to Question #116758 in General Chemistry for Nadeem

Question

Use pH = -log10[H+] to calculate the pH of each of the strong acid mystery solutions (Mystery HBr and Mystery H2SO4) based on the concentrations you determined in questions 4 and 5.

[H2SO4] = _____ pH H2SO4 = _____ [HBr] = _____ pH HBr = _____



Apply: The pH of a strong base is found with the equation pH = 14 + log10[OH-]. Based on their concentrations, find the pH of each of the strong bases.

[Ca(OH)2] = _____ pH Ca(OH)2 = _____ [NaOH] = _____ pH NaOH = _____







A. Make an equation for determining the number of moles of NaOH that are added to the flask based on [NaOH] and volume of NaOH titrant (mL NaOH):

Moles NaOH =

B. Make a similar expression for the number of moles of H2SO4 in the flask based on [H2SO4] and the volume of H2SO4 (mL).

Moles H2SO4 =

C. Because there are twice as many moles of NaOH as moles of H2SO4 in this reaction, you can say:
:

Answer 1

Assume that the concentrations of the acids were:

C[H₂SO₄] = 0.04 mol/l, then pH[H₂SO₄] = -log[0.08]=1.1;

C[HBr] = 0.001 mol/l, then pH[HBr] = -log[0.001]=3. 

Assume that the concentrations of the bases were:

C[Ca(OH)₂] = 0.025 mol/l, then pH[Ca(OH)₂]= 14+log[0.05]=12.7;

C[NaOH] = 0.00375 mol/l, then pH[NaOH]= 14+log[0.00375]=11.6.

A. The concentration is a number that shows us how many moles there are in the given volume:

$$C=\frac{n}{V},\rightarrow n=CV.$$

Therefore, for 0.00375 mol/l of NaOH the number of moles will be

$$n=0.00375\cdot V.$$

If we had 500 mL (it is 0.5 L) of titrant:

$$n[\text{NaOH}]=0.00375\cdot0.5=1.875\cdot10^{-3}\text{ mol}.$$

B. Similar to the previous problem, if we had 300 ml of the acid:

$$n[\text{H}_2\text{SO}_4]=0.04\cdot0.3=0.012\text{ mol}.$$

C. Because there are twice as many moles of NaOH as moles of H₂SO₄ in this reaction, we can say that there are two times more molecules of NaOH than molecules of H₂SO₄.