Answer:
pH = 3.12
Explanation:
To solve the problem you need to know that the pH is equal to the logarithm of the concentration of hydrogen ions, with negative sign. Or in other words pH is defined by:
![pH=-log[H^{+}]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%7B%2B%7D%5D)
As the problem gives you the concentration of hydrogen ions, you should replace the value:
![pH=-log[7.5*10^{-4}]](https://tex.z-dn.net/?f=pH%3D-log%5B7.5%2A10%5E%7B-4%7D%5D)
metal- a solid material that is typically hard, shiny, malleable, fusible, and ductile, with good electrical and thermal conductivity
non-metals- an element or substance that is not a metal.
metaloids- an element properties are intermediate between those of metals and solid nonmetals or semiconductors.
substance- a particular kind of matter with uniform properties.
physical properties- any property that is measurable, whose value describes a state of a physical system
luster- a gentle sheen or soft glow, especially that of a partly reflective surface.
conductivity- the degree to which a specified material conducts electricity, calculated as the ratio of the current density in the material to the electric field that causes the flow of current. It is the reciprocal of the resistivity.
Answer:
The solution becomes more concentrated
Explanation:
As the solvent evaporates, the solute remains intact. This implies that we have more of the solute than the solvent in the solution. So as the evaporation continues, the resulting solution becomes more concentrated as we now have more of the solute than the solvent.
I think answer should be carbon dioxide
6.8 is the pH of the solution after 10 ml of 5M NaOH is added.
Explanation:
Data given:
Molarity of C6H5CCOH = 0.100 M
molarity of ca(c6h5coo)2 = 0.2 M
Ka = 6.3 x 10^-5
first pH is calculated of the buffer solution
pH = pKa+ log 10 ![\frac{[A-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA-%5D%7D%7B%5BHA%5D%7D)
pKa = -log10[Ka]
pka = -log[6.3 x10^-5]
pKa = 4.200
putting the values to know pH of the buffer
pH = 4.200 + log 10 
pH = 4.200 + 0.3
pH = 4.5 (when NaOH was not added, this is pH of buffer solution)
now the molarity of the solution is calculated after NaOH i.e Mbuffer is added
MbufferVbuffer = Mbase Vbase
putting the values in above equation:
Mbuffer = 
= 
= 0.01 M
molarity or [ A-] = 5M
pH = pKa+ log 10
pH = 4.200 + log 10 
pH = 4.200+ 2.69
pH = 6.8
