When Ka =1.37 x 10^-4 
∴Pka = - ㏒Ka
          = -㏒ 1.37 x 10^-4 
          
           = 3.86
According to H- H equation: 
when PH = PKA + ㏒[conjugate base / weak acid]
when the lactate is the conjugate base and lactic acid is the weak acid 
∴ PH = Pka +㏒[lactate] / [lactic acid]
so, by substitution:
4.49 = 3.86 + ㏒[lactate]/[lactic acid]
∴[lactate]/[lactic acid] = 4.3
∴ [lactic acid] / [lactate] = 1/4.3 
            
                                        = 0.23
        
             
        
        
        
Answer:
D. Na2O
Because we already know that sodium oxide is a binary ionic compound because it contains one metal cation and one non-metal anion. 
 
        
                    
             
        
        
        
For each ionic compound formula, identify the main group to group 17  (VIIA).
Group17 (VIIA) is the primary group to which X belongs in each ionic combination with the formula CaX2. X is undoubtedly an electron acceptor. It is capable of taking an electron. We determine that X must be in the oxidation state -1 by looking at the chemical formula. The neutral ionic compound's total number of oxidation states is zero. Since calcium has an oxidation state of +2, we must identify the element that contains an anion with an oxidation state of -1. The halogens have a strong propensity to pick up an electron and create an anion. X belongs to group17 (VIIA) as a result.
To know more about ionic compound formula refer the link:
brainly.com/question/9597623
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C is Carbon
H is Hydrogen
O is Oxygen
The answer is B, D and E
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