<span>Which of the following is an electronic device that measures pH?
</span><span>a. a pH meter</span>
<span>THE HIGHEST CONCENTRATION OF HYDROGEN IONS IS LOCATED IN THE INTER-MEMBRANE SPACE. HYDROGEN IONS REACH THE INTER-MEMBRANE SPACE THROUGH PROTEIN CHANNELS EMBEDDED IN THE MITOCHONDRIAL MEMBRANE. THE MAIN FUNCTION OF INTER MEMBRANEIS OXIDATIVE PHOSPHORLATON. ENERGY IS REQUIRED TO MOVE THE HYDROGEN IONS ACROSS THE MEMBRANE BECAUSE THE HYDROGEN IONS ARE MOVING AGAINST THE CONCENTRATION GRADIENT. H+ GOES AGAINST THE CONCENTRATION GRADIENT THE USE OF THE GRADIENT TO DRIVE ATP SYNTHASE. HYDOGEN IONS DRIVE ATP SYNTHASE IN PHTOSYNTHESIS. THIS HAPPENS WHEN HYDROGEN IONS GET PUSHED ACROSS THE MEMBRANE CREATING A HIGH CONCENTRATION INSIDE THE THYLAKOID.</span>
The alkali metals, which occupy group 1 of the periodic table. This is because the valence shells of these elements have only 1 electron, so easily form an ionic bond with a non-metal compound by donating this. A cation is formed by this donation, since there is one fewer electron orbiting the nucleus than there is in the atomic form - conversely an anion is formed when an atom gains an extra electron to become negatively charged.
The ph level depends on temperature.
Part A
75.0 mL of 0.10 M HF; 55.0 mL of 0.15 M NaF
This combination will form a buffer.
Explanation
Here, weak acid HF and its conjugate base F- is available in the solution
Part B
150.0 mL of 0.10 M HF; 135.0 mL of 0.175 M HCl
This combination cannot form a buffer.
Explanation
Here, moles of HF = 0.15 x 0.1 = 0.015 moles
Moles of HCl = 0.135 x 0.175 = 0.023
Since HCl is a strong acid and the number of HCl is higher than HF. This prevents the dissociation of HF and the conjugate base F- will not be available in the solution
Part C
165.0 mL of 0.10 M HF; 135.0 mL of 0.050 M KOH
This combination will form a buffer.
Explanation
Moles of HF = 0.165 x 0.1 = 0.0165 moles
Moles of KOH = 0.135 x 0.05 = 0.00675 moles
Moles of KOH is not sufficient for the complete neutralization of HF. Thus weak acid HF and its conjugate base F- is available in the solution and form a buffer
Part D
125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 M CH3NH3Cl
This combination will form a buffer
Explanation
Here, weak acid CH3NH3+ and its conjugate base CH3NH2 is available in the solution and form a buffer
Part E
105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl
This combination will form a buffer
Explanation
Moles of CH3NH2 = 0.105 x 0.15 = 0.01575 moles
Moles of HCl = 0.095 x 0.1 = 0.0095 moles
Thus the HCl completely reacts with CH3NH2 and converts a part of the CH3NH2 to CH3NH3+. This results weak acid CH3NH3+ and its conjugate base CH3NH2 is in the solution and form a buffer
